BeerVectorDynamicsISM_Ch11_v2

CHAPTER 11

PROBLEM 11.1 A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS coordinates are used to determine his displacement as a function of time: x= 0.5t3 + t2 + 2t where x and t are expressed in ft and seconds, respectively. Determine the position, velocity, and acceleration of the boarder when t = 5 seconds.

SOLUTION Position:

Velocity: Acceleration: At t  5 s,

x  0.5t 3  t 2  2t

v

dx  1.5t 2  2t  2 dt

a

dv  3t  2 dt

x  0.5  5  52  2  5

x  97.5 ft 

v  1.5 5  2  5  2

v  49.5 ft/s 

3

2

a  3  5  2 

a  17 ft/s 2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.2 The motion of a particle is defined by the relation x  2t 3  9t 2  12t  10, where x and t are expressed in feet and seconds, respectively. Determine the time, the position, and the acceleration of the particle when v  0.

SOLUTION x  2t 3  9t 2  12t  10

Differentiating,

v

dx  6t 2  18t  12  6(t 2  3t  2) dt  6(t  2)(t  1)

a

dv  12t  18 dt

So v  0 at t  1 s and t  2 s. At t  1 s,

x1  2  9  12  10  15 a1  12  18  6

t  1.000 s 

x1  15.00 ft 

a1  6.00 ft/s2  At t  2 s,

x2  2(2)3  9(2)2  12(2)  10  14

t  2.00 s 

x2  14.00 ft  a2  (12)(2)  18  6

a2  6.00 ft/s 2 

PROBLEM 11.3 The vertical motion of mass A is defined by the relation x  10 sin 2t  15cos 2t  100, where x and t are expressed in mm and seconds, respectively. Determine (a) the position, velocity and acceleration of A when t  1 s, (b) the maximum velocity and acceleration of A.

SOLUTION x  10sin 2t  15cos 2t  100

v

dx  20 cos 2t  30sin 2t dt

a

dv  40sin 2t  60cos 2t dt

For trigonometric functions set calculator to radians: (a) At t  1 s.

x1  10sin 2  15cos 2  100  102.9 v1  20 cos 2  30sin 2  35.6 a1  40sin 2  60cos 2  11.40

x1  102.9 mm  v1  35.6 mm/s 

a1  11.40 mm/s2 

(b) Maximum velocity occurs when a  0. 40sin 2t  60 cos 2t  0

tan 2t  

60  1.5 40

2t  tan 1 (1.5)  0.9828 and 0.9828   Reject the negative value. 2t  2.1588

t  1.0794 s t  1.0794 s for vmax so

vmax  20cos(2.1588)  30sin(2.1588)  36.056

vmax  36.1 mm/s 

Note that we could have also used

vmax  202  302  36.056 by combining the sine and cosine terms. For amax we can take the derivative and set equal to zero or just combine the sine and cosine terms.

amax  402  602  72.1 mm/s 2

amax  72.1 mm/s2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.4 A loaded railroad car is rolling at a constant velocity when it couples with a spring and dashpot bumper system. After the coupling, the motion of the car is defined by the relation x  60e4.8t sin16t where x and t are expressed in mm and seconds, respectively. Determine the position, the velocity and the acceleration of the railroad car when (a) t  0, (b) t  0.3 s.

SOLUTION x  60e4.8t sin16t dx  60(4.8)e 4.8t sin16t  60(16)e 4.8t cos16t dt v  288e 4.8t sin16t  960e 4.8t cos16t v

a

dv  1382.4e4.8t sin16t  4608e4.8t cos16t dt  4608e4.8t cos16t  15360e4.8t sin16t

a  13977.6e4.8t sin16t  9216e4.8 cos16t (a) At t  0,

x0  0 v0  960 mm/s

a0  9216 mm/s 2 (b) At t  0.3 s,

x0  0 mm  v0  960 mm/s

a0  9220 mm/s2

 

e 4.8t  e 1.44  0.23692 sin16t  sin 4.8  0.99616 cos16t  cos 4.8  0.08750

x0.3  (60)(0.23692)(0.99616)  14.16

x0.3  14.16 mm



v0.3  87.9 mm/s



v0.3  (288)(0.23692)(0.99616)  (960)(0.23692)(0.08750)  87.9

a0.3  (13977.6)(0.23692)(0.99616)  (9216)(0.23692)(0.08750)  3108

a0.3  3110 mm/s 2  or 3.11 m/s 2



PROBLEM 11.5 The motion of a particle is defined by the relation x  6t 4  2t 3  12t 2  3t  3, where x and t are expressed in meters and seconds, respectively. Determine the time, the position, and the velocity when a  0.

SOLUTION We have

x  6t 4  2t 3  12t 2  3t  3

Then

v

dx  24t 3  6t 2  24t  3 dt

and

a

dv  72t 2  12t  24 dt

When a  0:

72t 2  12t  24  12(6t 2  t  2)  0 (3t  2)(2t  1)  0

or t

or At t 

2 s: 3

2 1 s and t   s (Reject) 3 2 4

3

2

2 2 2 2 x2/3  6    2    12    3    3 3 3 3 3 3

v2/3

t  0.667 s 

or

x2/3  0.259 m 

2

2 2 2  24    6    24    3 3 3 3

or v2/3  8.56 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.6 The motion of a particle is defined by the relation x  t 3  9t 2  24t  8, where x and t are expressed in inches and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

SOLUTION We have

x  t 3  9t 2  24t  8

Then

v

dx  3t 2  18t  24 dt

and

a

dv  6 t  18 dt

(a)

When v  0:

3 t 2  18t  24  3(t 2  6t  8)  0 (t  2)(t  4)  0 t  2.00 s and t  4.00 s 

(b)

When a  0:

6t  18  0 or t  3 s

x3  (3)3  9(3)2  24(3)  8

At t  3 s: First observe that 0  t  2 s:

v0

2 s  t  3 s:

v0

or

x3  10.00 in. 

Now

At t  0:

x0  8 in.

At t  2 s:

x2  (2)3  9(2)2  24(2)  8  12 in.

Then

x2  x0  12  (8)  20 in. | x3  x2 |  |10  12|  2 in.

Total distance traveled  (20  2) in.

Total distance  22.0 in. 

PROBLEM 11.7 A girl operates a radio-controlled model car in a vacant parking lot. The girl’s position is at the origin of the xy coordinate axes, and the surface of the parking lot lies in the x-y plane. She drives the car in a straight line so that the x coordinate is defined by the relation x(t) = 0.5t3 - 3t2 + 3t + 2, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and total distance travelled when the acceleration is zero.

SOLUTION Position:

x  t   0.5t 3  3t 2  3t  2

Velocity:

v t  

dx dt

v  t   1.5t 2  6t  3 (a) Time when v=0

0  1.5t 2  6t  3

t Acceleration:

6  62  4 1.5  3

a t  

t  0.586 s and t  3.414 s 

2  1.5 dv dt

a  t   3t  6 Time when a=0

0  3t  6 t2s

(b) Position at t=2 s

x  2   0.5  2   3  2   3  2  2 3

2

x  2  0 m  To find total distance note that car changes direction at t=0.586 s Position at t=0 s

x  0   0.5  0   3  0   3  0  2 3

2

x  0  2 Position at t=0.586 s

x  0.586   0.5  0.586   3  0.586   3  0.586  2 3

2

x  0.586  2.828 m Distances traveled: From t=0 to t=0.586 s:

x  0.586   x  0   0.828 m

From t=0.586 to t=2 s:

x  2   x  0.586   2.828 m

Total distance traveled  0.828 m  2.828 m

Total distance  3.656 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.8 The motion of a particle is defined by the relation x  t   t  2 , where x and t are expressed in feet and seconds, respectively. Determine (a) the two positions at which the velocity is zero, (b) the total distance traveled by the particle from t  0 to t  4 s.. 3

2

SOLUTION Position:

x  t   t 2   t  2

Velocity:

v t  

3

dx dt

v  t   2t  3  t  2 

2



v  t   2t  3 t 2  4t  4



 3t 2  14t  12

Time when v(t)=0

0  3t 2  14t  12

t

14  142  4  3 12 

2  3 t  1.131 s and t  3.535 s

(a) Position at t=1.131 s

x 1.131  1.131  1.1.31  2  2

3

x 1.131  1.935 ft.  Position at t=3.535 s

x  3.535    3.535    3.535  2  2

3

x  3.531  8.879 ft.  To find total distance traveled note that the particle changes direction at t=1.131 s and again at t=3.535 s. Position at t=0 s

x 0  0  0  2 2

3

x  4   4   4  2

3

x  0  8 ft Position at t=4 s

2

x  4  8 ft (b) Distances traveled: From t=0 to t=1.131 s:

x 1.131  x  0   6.065 ft.

From t=1.131 to t=3.531 s:

x  3.535   x 1.131  6.944 ft.

From t=3.531 to t=4 s:

x  4   x  3.535   0.879 ft.

Total distance traveled  6.065 ft  6.944 ft  0.879 ft

Total distance  13.888 ft. 

PROBLEM 11.9 The brakes of a car are applied, causing it to slow down at a rate of 10 ft/s2. Knowing that the car stops in 100 ft, determine (a) how fast the car was traveling immediately before the brakes were applied, (b) the time required for the car to stop.

SOLUTION a  10 ft/s 2 (a)

Velocity at x  0.

v



0 v0

0

(b)

dv  a  10 dx vdv  



xf 0

(10)dx

v02  10 x f  (10)(300) 2 v02  6000

v0  77.5 ft/s 

Time to stop. dv  a  10 dx



0

dv  

v0



tf 0

10dt

0  v0  10t f tf 

v0 77.5  10 10

t f  7.75 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.10 The acceleration of a particle is defined by the relation a  3e  0.2 t , where a and t are expressed in ft/s 2 and seconds, respectively. Knowing that x 0 and v 0 at t 0, determine the velocity and position of the particle when t 0.5 s.

SOLUTION Acceleration:

a  3e0.2t ft/s2

Given :

v0  0 ft/s,

Velocity:

a

x0  0 ft

dv  dv  adt dt

v

t

v0

0

 dv   adt t

v  vo 

 0



t

3e 0.2t dt  v  0=  15 e0.2t

0



v  15 1  e0.2t ft/s

Position:

v x



x0

dx  dx  vdt dt t



dx  vdt 0

t

 





x  xo  15 1  e 0.2t dt  x  0=15 t  5 e 0.2t 0

 v  15 1  e

  ft/s



t 0

x  15 t  5 e 0.2t  75 ft

Velocity at t=0.5 s

Position at t=0.5 s



0.20.5



v  0.5  1.427 ft/s 

x  15 0.5  5 e 0.20.5  75 ft

x  0.5  0.363 ft. 

PROBLEM 11.11 The acceleration of a particle is directly proportional to the square of the time t. When t  0, the particle is at x  24 m. Knowing that at t  6 s, x  96 m and v  18 m/s, express x and v in terms of t.

SOLUTION a  kt 2

We have

k  constant

dv  a  kt 2 dt

Now v



or

1 v  18  k (t 3  216) 3

18

dv 



t

At t  6 s, v  18 m/s:

6

kt 2 dt

1 v  18  k (t 3  216)(m/s) 3

or

dx 1  v  18  k (t 3  216) dt 3

Also

x

1  18  k (t 3  216)  dt 3 



or

1 1  x  24  18t  k  t 4  216t  3 4 

24

dx 



t

At t  0, x  24 m:

0 

Now At t  6 s, x  96 m: or Then or and or

1 1  96  24  18(6)  k  (6) 4  216(6)  3 4  k

1 m/s 4 9

1  1  1  x  24  18t    t 4  216t  3  9  4  x(t ) 

1 4 t  10t  24 108



11 v  18    (t 3  216) 3 9  v(t ) 

1 3 t  10 27

Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 11.12 The acceleration of a particle is defined by the relation a  kt 2 . (a) Knowing that v  8 m/s when t  0 and that v  8 m/s when t  2 s, determine the constant k. (b) Write the equations of motion, knowing also that x  0 when t  2 s.

SOLUTION a  kt 2 dv  a  kt 2 dt

(1)

t  0, v  8 m/s and t  2 s, v  8 ft/s



(a)

8 8

dv 



2 0

kt 2 dt

1 8  (8)  k (2)3 3

(b)

k  6.00 m/s 4 

Substituting k  6 m/s4 into (1) dv  a  6t 2 dt

t  0, v  8 m/s:



v 8

dv 



t

0

a  6t 2 

6t 2 dt

1 v  (8)  6(t )3 3

v  2t 3  8 

dx  v  2t 3  8 dt t  2 s, x  0:



x 0

dx 



t 2

(2t 3  8) dt; x 

1 4 t  8t 2

1  1  x   t 4  8t    (2)4  8(2)  2  2  1 x  t 4  8t  8  16 2

t 2

x

1 4 t  8t  8  2

PROBLEM 11.13 A Scotch yoke is a mechanism that transforms the circular motion of a crank into the reciprocating motion of a shaft (or vice versa). It has been used in a number of different internal combustion engines and in control valves. In the Scotch yoke shown, the acceleration of Point A is defined by the relation a 1.8sin kt, where a and t are expressed in m/s2 and seconds, respectively, and k 3 rad/s. Knowing that x 0 and v 0.6 m/s when t 0, determine the velocity and position of Point A when t 0.5 s.

SOLUTION Acceleration:

a  1.8sin kt m/s2

Given :

v0  0.6 m/s, x0  0,

Velocity:

dv a  dv  adt  dt

k  3 rad/s v

t

v0

0

 dv   adt 1.8

t t  0 a dt  1.8  0 sin kt dt  k cos kt

v  v0  v  0.6 

t 0

1.8 (cos kt  1)  0.6 cos kt  0.6 3

v  0.6cos kt m/s Position:

v

x

dx  dx  vdt  dt t



t



dx  vdt

x0

0

t

x  x0   0 v dt  0.6  0 cos kt dt  x0

0.6 sin kt k

t 0

0.6 (sin kt  0)  0.2sin kt 3

x  0.2sin kt m

When t =0.5 s,

kt  (3)(0.5)  1.5 rad

v  0.6cos1.5  0.0424 m/s

v  42.4 mm/s 

x  0.2sin1.5  0.1995 m

x  199.5 mm 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.14 For the scotch yoke mechanism shown, the acceleration of Point A is defined by the relation a  1.08sin kt  1.44 cos kt , where a and t are expressed in m/s2 and seconds, respectively, and k = 3 rad/s. Knowing that x = 0.16 m and v = 0.36 m/s when t = 0, determine the velocity and position of Point A when t = 0.5 s.

SOLUTION Acceleration:

a  1.08sin kt  1.44cos kt m/s2

Given :

v0  0.36 m/s,

Velocity:

dv a  dv  adt  dt

Integrate:

x0  0.16,

k  3 rad/s

v

t

v0

0

 dv   adt

t

t

v  v0  1.08  0 sin kt dt  1.44  0 cos kt dt v  0.36  

1.08 cos kt k

t 0



1.44 sin kt k

t 0

1.08 1.44 (cos 3t  1)  (sin 3t  0) 3 3

 0.36 cos 3t  0.36  0.48sin 3t v  0.36 cos 3t  0.48sin 3t m/s Evaluate at t  0.5 s

 0.36 cos1.5  0.36  0.48sin1.5

Position:

v

dx  dx  vdt  dt

x



t



dx  vdt 0

x0

t

v  453 mm/s 

t

t

x  x0   0 v dt  0.36  0 cos kt dt  0.48  0 sin kt dt x  0.16 

0.36 sin kt k

t 0



0.48 cos kt k

t 0

0.36 0.48 (sin 3t  0)  (cos 3t  1) 3 3  0.12sin 3t  0.16 cos 3t  0.16 

x  0.12sin 3t  0.16 cos 3t m

Evaluate at t  0.5 s

x  0.12sin1.5  0.16 cos1.5  0.1310 m

x  131.0 mm 

PROBLEM 11.15 A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After contact the equipment experiences an acceleration of a   kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.

SOLUTION a

vdv  k x dx

Separate and integrate.



vf v0

vdv  



xf 0

k x dx

1 2 1 2 1 v f  v0   kx 2 2 2 2

xf

0

1   k x 2f 2

Use v0  4 m/s, x f  0.02 m, and v f  0. Solve for k.

1 1 0  (4)2   k (0.02)2 2 2

k  40, 000 s 2

Maximum acceleration.

amax  kxmax : (40,000)(0.02)  800 m/s 2 a  800 m/s 2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.16 A projectile enters a resisting medium at x  0 with an initial velocity v0  900 ft/s and travels 4 in. before coming to rest. Assuming that the velocity of the projectile is defined by the relation v  v0  kx, where v is expressed in ft/s and x is in feet, determine (a) the initial acceleration of the projectile, (b) the time required for the projectile to penetrate 3.9 in. into the resisting medium.

SOLUTION First note When x 

 4  0  (900 ft/s)  k  ft   12 

4 ft, v  0: 12

k  2700

or (a)

1 s

We have

v  v0  kx

Then

a

or

a   k (v0  kx)

At t  0:

1 a  2700 (900 ft/s  0) s

dv d  (v0  kx)   kv dt dt

a0  2.43  106 ft/s 2 

or (b)

dx  v  v0  kx dt

We have At t  0, x  0: or



x

0

dx  v0  kx

t

 dt 0

1  [ln(v0  kx)]0x  t k

or

t

1  v0  1  1   ln    ln  k  v0  kx  k  1  vk x  0  

When x  3.9 in.:

t

1   1 ln   2700 1/s 3.9 2700 1s 1  900 ft/s  12 ft  

or

t  1.366  103 s 

PROBLEM 11.17 The acceleration of a particle is defined by the relation a  k/x. It has been experimentally determined that v  15 ft/s when x  0.6 ft and that v  9 ft/s when x  1.2 ft. Determine (a) the velocity of the particle when x  1.5 ft, (b) the position of the particle at which its velocity is zero.

SOLUTION a

vdv  k  dx x

Separate and integrate using x  0.6 ft, v  15 ft/s.



v

15

vdv   k



x

0.6

v

dx x

1 2 v   k ln x 2 15

x

0.6

1 2 1  x  v  (15) 2   k ln   2 2  0.6 

(1)

When v  9 ft/s, x  1.2 ft 1 2 1  1.2  (9)  (15) 2  k ln   2 2  0.6  Solve for k. k  103.874 ft 2 /s 2 (a)

Velocity when x  65 ft. Substitute

k  103.874 ft 2 /s 2

and x  1.5 ft into (1).

1 2 1  1.5  v  (15) 2  103.874 ln   2 2  0.6  v  5.89 ft/s 

(b)

Position when for v  0,

1  x  0  (15) 2  103.874 ln   2  0.6   x  ln    1.083  0.6  



 x  1.772 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.18 A brass (nonmagnetic) block A and a steel magnet B are in equilibrium in a brass tube under the magnetic repelling force of another steel magnet C located at a distance x  0.004 m from B. The force is inversely proportional to the square of the distance between B and C. If block A is suddenly removed, the acceleration of block B is a  9.81  k /x 2 , where a and x are expressed in m/s2 and m, respectively, and k  4  104 m3 /s 2 . Determine the maximum velocity and acceleration of B.

SOLUTION 0  9.81 

The maximum velocity occurs when a  0. xm2 

k 4  104   40.775  106 m 2 9.81 9.81

k xm2

xm  0.0063855 m

The acceleration is given as a function of x. v

dv k  a  9.81  2 dx x

Separate variables and integrate:

vdv  9.81dx 



v 0



vdv  9.81

x

k dx x2

dx  k

x0



x x0

dx x2

1 1  1 2 v  9.81( x  x0 )  k    2  x x0   1 1 2 1    vm  9.81( xm  x0 )  k  2  xm x0  1 1    9.81(0.0063855  0.004)  (4  104 )     0.0063855 0.004   0.023402  0.037358  0.013956 m 2 /s 2

Maximum velocity:

vm  0.1671 m/s

vm  167.1 mm/s 

The maximum acceleration occurs when x is smallest, that is, x  0.004 m.

am  9.81 

4  104 (0.004)2

am  15.19 m/s 2



PROBLEM 11.19 Based on experimental observations, the acceleration of a particle is defined by the relation a  (0.1  sin x/b), where a and x are expressed in m/s2 and meters, respectively. Knowing that b  0.8 m and that v  1 m/s when x  0, determine (a) the velocity of the particle when x  1 m, (b) the position where the velocity is maximum, (c) the maximum velocity.

SOLUTION We have When x  0, v  1 m/s:

v



v 1

dv x    a    0.1  sin  0.8  dx 

vdv 



x     0.1  sin  dx 0  0.8  x

x

1 2 x   (v  1)   0.1x  0.8 cos 2 0.8  0 

or

x 1 2 v  0.1x  0.8 cos  0.3 2 0.8

or (a)

When x  1 m:

1 2 1 v  0.1(1)  0.8 cos  0.3 2 0.8

v   0.323 m/s 

or (b)

When v  vmax, a  0: or

(c)

x     0.1  sin 0 0.8  

x  0.080 134 m

x  0.0801 m 

When x  0.080 134 m: 1 2 0.080 134 vmax  0.1(0.080 134)  0.8 cos  0.3 2 0.8  0.504 m 2 /s 2 or

vmax  1.004 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.20 A spring AB is attached to a support at A and to a collar. The unstretched length of the spring is l. Knowing that the collar is released from rest at x  x0 and has an acceleration defined by the relation a  100( x  lx / l 2  x 2 ) , determine the velocity of the collar as it passes through Point C.

SOLUTION av

Since a is function of x,

 dv lx  100  x   dx l 2  x2 

   

Separate variables and integrate:



vf v0

vdv  100



 lx x 2 x0  l  x2  0

 x2 1 2 1 2 v f  v0  100   l l 2  x2 2 2  2

  dx  

  

0

x0

 x2 1 2 v f  0  100   0  l 2  l l 2  x02  2 2 

  

1 2 100 2 (l  x02  l 2  2l l 2  x02 ) vf  2 2 100 ( l 2  x02  l ) 2  2 v f  10( l 2  x02  l ) 

PROBLEM 11.21





The acceleration of a particle is defined by the relation a  k 1  e  x , where k is a constant. Knowing that the velocity of the particle is v   9 m/s when x   3 m and that the particle comes to rest at the origin, determine (a) the value of k, (b) the velocity of the particle when x   2 m.

SOLUTION





Acceleration:

a  k 1  e x

Given :

at x   3 m, v  9 m/s at x  0 m, v  0 m/s adx  vdv



k 1 e

x

 dx  vdv

Integrate using x=-3 m and v=9 m/s as the lower limits of the integrals x

v

 k 1  e  dx   vdv x

3

9



k x  e x

Velocity:





x

v

 3

1 2 v 2 9



k x  e  x  3  e3

  12 v

2

1  (9) 2 2

(1)

(a) Now substitute v=0 m/s and x=0 m into (1) and solve for k





k 0  e 0  3  e3

  12 0

2

1  (9)2 2

k  2.52 m 2 /s 2  (b) Find velocity when x= -2 m using the equation (1) and the value of k





2.518 2  e 2  3  e3

  12 v

2

1  (9) 2 2 v  4.70 m/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.22 2 Starting from x  0 with no initial velocity, a particle is given an acceleration a  0.1 v  16, where a 2 and v are expressed in ft/s and ft/s, respectively. Determine (a) the position of the particle when v  3ft/s, (b) the speed and acceleration of the particle when x  4 ft.

SOLUTION a

vdv  0.1(v 2  16)1/2 dx

(1)

Separate and integrate.



v 0

vdv v 2  16





x 0

0.1 dx

v

(v 2  16)1/2  0.1 x 0

2

1/2

(v  16)

(a)

(b)

 4  0.1 x x  10[(v 2  16)1/2  4]

(2)

x  10[(32  16)1/2  4]

x  10.00 ft 

v  3 ft/s.

x  4 ft. From (2),

(v 2  16)1/2  4  0.1x  4  (0.1)(4)  4.4 v 2  16  19.36 v 2  3.36ft 2 /s2

From (1),

a  0.1(1.8332  16)1/2

v  1.833 ft/s 

a  0.440 ft/s 2 

PROBLEM 11.23 A ball is dropped from a boat so that it strikes the surface of a lake with a speed of 16.5 ft/s. While in the water the ball experiences an acceleration of a  10  0.8v, where a and v are expressed in ft/s2 and ft/s, respectively. Knowing the ball takes 3 s to reach the bottom of the lake, determine (a) the depth of the lake, (b) the speed of the ball when it hits the bottom of the lake.

SOLUTION a

dv  10  0.8v dt

Separate and integrate:



dv  v0 10  0.8v v



t 0

dt

v



1 ln(10  0.8v)  t 0.8 v0  10  0.8v ln   10  0.8v0

   0.8t 

10  0.8v  (10  0.8v0 )e0.8t 0.8v  10  (10  0.8v0 )e0.8t

or

v  12.5  (12.5  v0 )e0.8t v  12.5  4e0.8t

With v0  16.5ft/s

Integrate to determine x as a function of t. dx  12.5  4e0.8t v dt



x 0

dx 



t 0

(12.5  4e 0.8t )dt

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.23 (Continued) t

x  12.5t  5e0.8t  12.5t  5e0.8t  5 0

(a)

At t  35 s,

x  12.5(3)  5e2.4  5  42.046 ft (b)

v  12.5  4e2.4  12.863 ft/s

x  42.0 ft  v  12.86 ft/s 

PROBLEM 11.24 The acceleration of a particle is defined by the relation a  k v , where k is a constant. Knowing that x  0 and v  81 m/s at t  0 and that v  36 m/s when x  18 m, determine (a) the velocity of the particle when x  20 m, (b) the time required for the particle to come to rest.

SOLUTION (a)

We have

dv  a  k v dx

v

v dv  k dx

so that v



or

2 3/2 v [v ]81  kx 3

or

2 3/2 [v  729]  kx 3

When x  18 m, v  36 m/s:

v dv 

81



x

When x  0, v  81 m/s:

0

k dx

2 (363/2  729)  k (18) 3

k  19 m/s 2

or Finally When x  20 m:

2 3/2 (v  729)  19(20) 3

v3/2  159

or (b)

We have

dv  a  19 v dt

At t  0, v  81 m/s:



v

dv

81

v





t 0

19dt

or

v 2[ v ]81  19t

or

2( v  9)  19t

When v  0: or

v  29.3 m/s 

2(9)  19t t  0.947 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.25 The acceleration of a particle is defined by the relation a  kv2.5, where k is a constant. The particle starts at x  0 with a velocity of 16 mm/s, and when x  6 mm the velocity is observed to be 4 mm/s. Determine (a) the velocity of the particle when x  5 mm, (b) the time at which the velocity of the particle is 9 mm/s.

SOLUTION Acceleration:

a  kv 2.5

Given :

at t=0, x  0 mm, v  16 mm/s at x  6 mm, v  4 mm/s adx  vdv 2.5

kv dx  vdv Separate variables

 kdx  v  3 2 dv

Integrate using x=-0 m and v=16 mm/s as the lower limits of the integrals x



v



-kdx  v 3 2 dv

0

16 x

 2v 1 2

 kx 0

Velocity and position:

kx  2v 1 2 

v 16

1 2

(1)

Now substitute v=4 mm/s and x=6 mm into (1) and solve for k 1 2 k  0.0833 mm 3 2 s1 2 k  6   41 2 

(a) Find velocity when x= 5 mm using the equation (1) and the value of k k  5   2v 1 2 

1 2 v  4.76 mm/s

(b)

Separate variables

a

dv or adt  dv dt

a

dv or adt  dv dt

 kdt  v 2.5 dv

PROBLEM 11.25 (Continued) Integrate using t=0 and v=16 mm/s as the lower limits of the integrals t

v

 -kdt   v 0

v

kt 0

kt 

dv

16

t

Velocity and time:

5 2

2   v 3 2 3 16

2 3 2 1  v 3 96

(2)

Find time when v= 9 mm/s using the equation (2) and the value of k kt 

2 3 2 1  9  3 96 t  0.171 s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

Problem 11.26 A human powered vehicle (HPV) team wants to model the acceleration during the 260 m sprint race (the first 60 m is called a flying start) using a = A – Cv2, where a is acceleration in m/s2 and v is the velocity in m/s. From wind tunnel testing, they found that C = 0.0012 m-1. Knowing that the cyclist is going 100 km/h at the 260 meter mark, what is the value of A?

SOLUTION Acceleration:

a  A  Cv2 m/s2

Given :

C  0.0012 m 1 , v f  100 km/hr when x f  260 m Note: 100 km/hr = 27.78 m/s adx  vdv

 A  Cv  dx  vdv 2

Separate variables

dx 

vdv A  Cv 2

Integrate starting from rest and traveling a distance xf with a final velocity vf. xf

vf

0

0

vdv

 dx   A  Cv x

x 0f  

2



1 ln A  Cv 2 2C



1 ln A  v 2f xf   2C 1  A ln  xf  2C  A  Cv 2f Next solve for A

A

Cv 2f e 1 e



vf 0

  21C ln  A    

2 Cx f

2 Cx f

Now substitute values of C, xf and vf and solve for A

A  1.995 m/s 2 

PROBLEM 11.27 Experimental data indicate that in a region downstream of a given louvered supply vent the velocity of the emitted air is defined by v  0.18v0 /x, where v and x are expressed in m/s and meters, respectively, and v0 is the initial discharge velocity of the air. For v0  3.6 m/s, determine (a) the acceleration of the air at x  2 m, (b) the time required for the air to flow from x  1 to x  3 m.

SOLUTION (a)

dv dx 0.18v0 d  0.18v0   x dx  x 

av

We have



a

When x  2 m:

0.0324v02 x3

0.0324(3.6)2 (2)3 a  0.0525 m/s 2 

or (b)

0.18v0 dx v dt x

We have From x  1 m to x  3 m:



3 1

xdx 



t3 t1

0.18v0 dt

3

or

1 2  2 x   0.18v0 (t3  t1 )  1

or

(t3  t1 ) 

or

1 2

(9  1)

0.18(3.6) t3  t1  6.17 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.28 Based on observations, the speed of a jogger can be approximated by the relation v  7.5(1  0.04 x)0.3 , where v and x are expressed in mi/h and miles, respectively. Knowing that x  0 at t  0, determine (a) the distance the jogger has run when t  1 h, (b) the jogger’s acceleration in ft/s2 at t  0, (c) the time required for the jogger to run 6 mi.

SOLUTION (a)

dx  v  7.5(1  0.04 x)0.3 dt

We have



At t  0, x  0: or

x

0

dx  (1  0.04 x)0.3

t

 7.5dt 0

1  1  0.7 x   [(1  0.04 x) ]0  7.5t 0.7  0.04 

1  (1  0.04 x)0.7  0.21t

or or

x

1 [1  (1  0.21t )1/0.7 ] 0.04

At t  1 h:

x

1 {1  [1  0.21(1)]1/0.7 } 0.04

(1)

x  7.15 mi 

or (b)

We have

av

dv dx

d [7.5(1  0.04 x)0.3 ] dx  7.52 (1  0.04 x)0.3 [(0.3)(0.04)(1  0.04 x) 0.7 ]  7.5(1  0.04 x)0.3

 0.675(1  0.04 x)0.4 At t  0, x  0:

a0  0.675 mi/h 2 

5280 ft  1 h    1 mi  3600 s 

a0  275  106 ft/s 2 

or (c)

From Eq. (1)

t

When x  6 mi:

t

or

2

1 [1  (1  0.04 x)0.7 ] 0.21

1 {1  [1  0.04(6)]0.7 } 0.21  0.83229 h

t  49.9 min 

PROBLEM 11.29 The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as 32.2 [1  ( y/20.9  106 )]2

a

where a and y are expressed in ft/s2 and feet, respectively. Using this expression, compute the height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is (a) 1800 ft/s, (b) 3000 ft/s, (c) 36,700 ft/s.

SOLUTION We have

v

dv a dy

1 

32.2 y 20.9  106

When

y  0,

provided that v does reduce to zero,

y  ymax , v  0



Then

v0

v dv 



v  v0

32.2

ymax 0



1

y 20.9  106



2

dy

 1 v02  1345.96  106 1   1  ymax 20.9  106 

or

ymax 

or

v0  1800 ft/s:

ymax 

v0  3000 ft/s: or

y

 max   0

   

v02 64.4 

v02 20.9  106

(1800)2 64.4 

(1800)2 20.9  106

ymax  50.4  103 ft 

or (b)

2

 1 1  v02  32.2  20.9  106 y  2 1 20.9  106 

or

(a)

0



ymax 

(3000) 2 64.4 

(3000)2 20.9  106

ymax  140.7  103 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.29 (Continued)

(c)

v0  36, 700 ft/s:

ymax 

(36, 700) 2 64.4 

(36,700)2 20.9  106

 3.03  1010 ft

This solution is invalid since the velocity does not reduce to zero. The velocity 36,700 ft/s is above the escape velocity vR from the earth. For vR and above. ymax



PROBLEM 11.30 The acceleration due to gravity of a particle falling toward the earth is a   gR 2 /r 2, where r is the distance from the center of the earth to the particle, R is the radius of the earth, and g is the acceleration due to gravity at the surface of the earth. If R  3960 mi, calculate the escape velocity, that is, the minimum velocity with which a particle must be projected vertically upward from the surface of the earth if it is not to return to the earth. (Hint: v  0 for r  .)

SOLUTION We have

v

dv gR 2 a 2 dr r

r  R, v  ve

When

r  , v  0 0

gR 2 dr r2



or

1 1   ve2  gR 2   2  r R

or

ve  2 gR

ve

vdv 





then

R





1/ 2

5280 ft     2  32.2 ft/s 2  3960 mi   1 mi   or

ve  36.7  103 ft/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.31 The velocity of a particle is v  v0 [1  sin ( t/T )]. Knowing that the particle starts from the origin with an initial velocity v0 , determine (a) its position and its acceleration at t  3T , (b) its average velocity during the interval t  0 to t  T .

SOLUTION (a)

 dx   t   v  v0 1  sin    dt  T  

We have

At t  0, x  0:



x 0

dx 



   t  v0 1  sin    dt 0  T   t

t

 T  T   t   t  T  x  v0 t  cos     v0 t  cos      T 0 T      

At t  3T :

 T    3T x3T  v0 3T  cos    T 

a At t  3T : (b)

2T   T       v0  3T       

(1)

x3T  2.36 v0T 

dv d    t   t     v0 1  sin      v0 cos dt dt   T T  T   

a3T  v0

 T

cos

  3T

a3T 

T

 v0 T



Using Eq. (1) At t  0: At t  T : Now

T T  x0  v0  0  cos(0)    0    

 T  T xT  v0 T  cos    T  vave 

2T  T       v0  T     

xT  x0 0.363v0T  0  t T 0

   0.363v0T  vave  0.363v0 

Problem 11.32 An eccentric circular cam, which serves a similar function as the Scotch yoke mechanism in Problem 11.13, is used in conjunction with a flat face follower to control motion in pumps and in steam engine valves. Knowing that the eccentricity is denoted by e, the maximum range of the displacement of the follower is dmax, and the maximum velocity of the follower is vmax, determine the displacement, velocity, and acceleration of the follower.

SOLUTION Constraint:

y  r  e cos 

(1)

Differentiate:

y  e sin 

(2)

Differentiate again:

 y  esin   e2 cos

(3)

ymax occurs when cosθ =1 and ymin occurs when cosθ = -1 d max  ymax  ymin d max  r  e   r  e  d max  2e

e

dmax 2

(4) yr

Substitute (4) into (1) to get Position

d max cos   2

Max Velocity occurs when sin   1

vmax  e

 

vmax e

(5) y  vmax sin 

Substitute (5) into (2) to get velocity and assume cw rotation. Substitute (4) and (5) into (3)

 y  Acceleration:

 4v 2 d max  d  sin   max  2max 2 2  d max

  cos    y 

d max  2v 2  sin   max cos   2 d max

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

Problem 11.33 An airplane begins its take-off run at A with zero velocity and a constant acceleration a. Knowing that it becomes airborne 30 s later at B and that the distance AB is 900 m, determine (a) the acceleration a, (b) the take-off velocity v B .

SOLUTION Given :

x A  0 , xB  900 m , v A  0 , t  30 s

Uniform Acceleration:

xB  xA  vAt 

1 2 at 2

Substitute known values:

900 m  0  0 

1 2 a  30 s  2 a  2.0 m/s 2 

(a) Solve for acceleration Uniform Acceleration:

vB  v A  at

Substitute known values:

vB  0  2 m/s2  30 s 

(b) Velocity at takeoff (point B)





vB  60.0 m/s

PROBLEM 11.34 A motorist is traveling at 54 km/h when she observes that a traffic light 240 m ahead of her turns red. The traffic light is timed to stay red for 24 s. If the motorist wishes to pass the light without stopping just as it turns green again, determine (a) the required uniform deceleration of the car, (b) the speed of the car as it passes the light.

SOLUTION Uniformly accelerated motion: x0  0

(a)

x  x0  v0t 

v0  54 km/h  15 m/s

1 2 at 2

when t  24s, x  240 m: 240 m  0  (15 m/s)(24 s) 

1 a (24 s) 2 2

a  0.4167 m/s 2

(b)

a  0.417 m/s 2 

v  v0  a t

when t  24s: v  (15 m/s)  ( 0.4167 m/s)(24 s) v  5.00 m/s v  18.00 km/h

v  18.00 km/h 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

Problem 11.35 Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a 750-ft ramp at a high speed v0 and travels 540 ft in 6 s at constant deceleration before its speed is reduced to v0 / 2. Assuming the same constant deceleration, determine (a) the additional time required for the truck to stop, (b) the additional distance traveled by the truck.

SOLUTION Given :

xo  0 , x A  540 m , t A  6 s , vA 

Uniform Acceleration:

v A  vo  at A

Substitute known values:

1 vo  vo   a  6  2 a

1 vo , Lramp  750 ft 2

1 vo 12 1 2 at A 2

Uniform Acceleration:

xA  xo  vot A 

Substitute known values:

540  0  vo  6  

1 2 vo  6  24

vo  120 ft/s and a  10 ft/s 2 (a)

vB  vo  at B

Substitute known values:

0  120  10tB

Solve for tB

tB  12 s tB  t A  6.0 s 

Additional time to stop

1 2 atB 2

(b)

xB  xo  votB 

Substitute known values:

1 2 xB  0  120 12   10 12  2

Solve for xB

xB  720 ft

Additional time to stop

xB  x A  180.0 ft 

PROBLEM 11.36 A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16 s later. Knowing that the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude and assuming that g  32.2 ft/s 2 , determine (a) the speed v1 of the rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.

SOLUTION (a)

1 2 at 2

We have

y  y1  v1t 

At tland ,

y0

Then

0  89.6 ft  v1 (16 s) 

1 (32.2 ft/s 2 )(16 s) 2 2 v1  252 ft/s 

or (b)

We have

v 2  v12  2a( y  y1 )

At

y  ymax ,

Then

0  (252 ft/s)2  2(32.2 ft/s 2 )( ymax  89.6) ft

or

v0

y max  1076 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.37 A small package is released from rest at A and moves along the skate wheel conveyor ABCD. The package has a uniform acceleration of 4.8 m/s 2 as it moves down sections AB and CD, and its velocity is constant between B and C. If the velocity of the package at D is 7.2 m/s, determine (a) the distance d between C and D, (b) the time required for the package to reach D.

SOLUTION (a)

For A

B and C

D we have

v 2  v02  2a( x  x0 ) 2 vBC  0  2(4.8 m/s2 )(3  0) m

Then, at B

 28.8 m2 /s 2

(vBC  5.3666 m/s)

2 vD2  vBC  2aCD ( xD  xC )

and at D

d  xD  xC

(7.2 m/s)2  (28.8 m 2 /s 2 )  2(4.8 m/s 2 )d

or

d  2.40 m 

or (b)

For A

B and C

D we have v  v0  at

Then A

B

5.3666 m/s  0  (4.8 m/s2 )t AB t AB  1.118 04 s

or and C

7.2 m/s  5.3666 m/s  (4.8 m/s 2 )tCD

D

tCD  0.38196 s

or Now, for B

C, we have

xC  x B  v BC t BC

or

3 m  (5.3666 m/s)t BC

or

t BC  0.55901 s

Finally, or

t D  t AB  t BC  tCD  (1.11804  0.55901  0.38196) s t D  2.06 s 

PROBLEM 11.38 A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runs with constant velocity. If the sprinter’s time for the first 35 m is 5.4 s, determine (a) his acceleration, (b) his final velocity, (c) his time for the race.

SOLUTION 0  x  35 m, a  constant

Given:

35 m  x  100 m, v  constant At t  0, v  0 when

x  35 m, t  5.4 s

Find: (a)

a

(b)

v when x  100 m

(c)

t when x  100 m

(a)

We have At t  5.4 s: or

x  0  0t  35 m 

1 2 at 2

for

0  x  35 m

1 a(5.4 s)2 2

a  2.4005 m/s 2 a  2.40 m/s 2 

(b)

First note that v  vmax for 35 m  x  100 m. Now

(c)

v 2  0  2a( x  0)

for

0  x  35 m

When x  35 m:

2 vmax  2(2.4005 m/s 2 )(35 m)

or

vmax  12.9628 m/s

We have When x  100 m: or

x  x1  v0 (t  t1 )

vmax  12.96 m/s 

for

35 m  x  100 m

100 m  35 m  (12.9628 m/s)(t 2  5.4) s

t 2  10.41 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.39 Automobile A starts from O and accelerates at the constant rate of 0.75 m/s2. A short time later it is passed by bus B which is traveling in the opposite direction at a constant speed of 6 m/s. Knowing that bus B passes point O 20 s after automobile A started from there, determine when and where the vehicles passed each other.

SOLUTION Place origin at 0. Motion of auto.

 xA 0  0,  vA 0  0, x A   x A 0   v A 0 t 

aA  0.75 m/s2 1 1 a At 2  0  0     0.75  t 2 2 2

x A  0.375t 2 m Motion of bus.

 xB 0  ?,  vB 0   6 m/s,

aB  0

xB   xB 0   vB 0 t   xB 0  6t m At t  20 s , x B  0.

0   xB 0   6  20  Hence,

 xB 0  120 m

x B  120  6 t

When the vehicles pass each other, xB  x A .

120  6t  0.375 t 2 0.375 t 2  6 t  120  0 t

t

6 

(6) 2   4  0.375   120 

 2  0.375 

 6  14.697  11.596 s 0.75

and  27.6 s t  11.60 s 

Reject the negative root. Corresponding values of xA and xB.

xA   0.37511.596  50.4 m 2

xB  120   6 11.596   50.4 m

x  50.4 m 

PROBLEM 11.40 In a boat race, boat A is leading boat B by 50 m and both boats are traveling at a constant speed of 180 km/h. At t  0, the boats accelerate at constant rates. Knowing that when B passes A, t  8 s and v A  225 km/h, determine (a) the acceleration of A, (b) the acceleration of B.

SOLUTION (a)

We have

v A  (v A )0  a At (v A ) 0  180 km/h  50 m/s

At t  8 s: Then

v A  225 km/h  62.5 m/s 62.5 m/s  50 m/s  a A (8 s)

a A  1.563 m/s2 

or (b)

We have

1 1 x A  ( x A )0  (v A )0 t  a At 2  50 m  (50 m/s)(8 s)  (1.5625 m/s2 )(8 s)2  500 m 2 2 and

1 xB  0  (vB ) 0 t  a B t 2 2

At t  8 s:

x A  xB

( v B ) 0  50 m/s

1 500 m  (50 m/s)(8 s)  aB (8 s)2 2 or

aB  3.13 m/s2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.41 As relay runner A enters the 65-ft-long exchange zone with a speed of 30 ft/s, he begins to slow down. He hands the baton to runner B 2.5 s later as they leave the exchange zone with the same velocity. Determine (a) the uniform acceleration of each of the runners, (b) when runner B should begin to run.

SOLUTION Let x  0 at the start of the exchange zone, and t  0 as runner A enters the exchange zone. Then,

 v A 0

Motion of runner A:

v A   v A  0  a At

 30 ft/s.

x A   v A 0 t 

1 a At 2 2

(a) Solving for a A , and noting that x A  65 ft when t  2.5 s aA 

2  x A   v A 0 t  t

2



2  65   30  2.5  

a A  3.20 ft/s 2 

2.52

 vA  f

At t  2.5 s,

 30   3.20  2.5  22.0 ft/s

Motion of runner B:  vB 0  0 and xB  0 at the starting time tB of runner B.

vB2   vB 0  2aB xB 2

Then,

Solving for aB , and noting that vB   vA  f  22.0 ft/s When xB  65 ft,

aB 

 vB 2   vB 02 2 xB



22.02  0  2  65

aB  3.723 ft/s2 

v B   v B 0  a B  t  t B   a B  t  t B 

t  tB 

vB aB

(b) Solving for t B , and noting that vB  22.0 ft/s when t  2.5 s tB  t 

vB 22.0  2.5   3.41 s aB 3.723

Runner B should begin 3.41 s before runner A reaches the exchange zone. 

PROBLEM 11.42 Automobiles A and B are traveling in adjacent highway lanes and at t  0 have the positions and speeds shown. Knowing that automobile A has a constant acceleration of 1.8 ft/s 2 and that B has a constant deceleration of 1.2 ft/s2 , determine (a) when and where A will overtake B, (b) the speed of each automobile at that time.

SOLUTION a A  1.8 ft/s2

aB  1.2 ft/s2

(v A )0  24 mi/h  35.2 ft/s

A overtakes B

(vB )0  36 mi/h  52.8 ft/s

Motion of auto A: v A  ( v A ) 0  a A t  35.2  1.8t

x A  ( x A ) 0  (v A ) 0 t 

(1)

1 1 a At 2  0  35.2t  (1.8)t 2 2 2

(2)

Motion of auto B: v B  ( v B ) 0  a B t  52.8  1.2t

xB  ( x B ) 0  ( vB ) 0 t  (a)

(3)

1 1 aB t 2  75  52.8t  (1.2)t 2 2 2

(4)

A overtakes B at t  t1 . x A  xB : 35.2t  0.9t12  75  52.8t1  0.6t12 1.5t12  17.6t1  75  0 t1  3.22 s

Eq. (2):

and

t1  15.0546

x A  35.2(15.05)  0.9(15.05)2

t1  15.05 s  x A  734 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.42 (Continued)

(b)

Velocities when t1  15.05 s Eq. (1):

v A  35.2  1.8(15.05)

v A  62.29 ft/s

Eq. (3):

v A  42.5 mi/h



vB  23.7 mi/h



vB  52.8  1.2(15.05) v B  34.74 ft/s

PROBLEM 11.43 Two automobiles A and B are approaching each other in adjacent highway lanes. At t  0, A and B are 3200 ft apart, their speeds are v A  65 mi/h and v B  40 mi/h, and they are at Points P and Q, respectively. Knowing that A passes Point Q 40 s after B was there and that B passes Point P 42 s after A was there, determine (a) the uniform accelerations of A and B, (b) when the vehicles pass each other, (c) the speed of B at that time.

SOLUTION (a)

x A  0  (v A ) 0 t 

We have (x is positive

3200 m  (95.333 m/s)(40 s) 

Also, xB  0  (vB )0 t 

1 aB t 2 2

Then (95.333 ft/s)t AB 

(c)

1 aB (42 s) 2 2

aB  0.83447 ft/s 2

When the cars pass each other

Solving

a A  0.767 ft/s2 

( v B ) 0  40 mi/h  58.667 ft/s

3200 ft  (58.667 ft/s)(42 s) 

or

or

1 a A (40 s) 2 2

; origin at Q.)

At t  42 s:

(b)

(v A )0  65 mi/h  95.33 ft/s

; origin at P.)

At t  40 s:

(xB is positive

1 a At 2 2

aB  0.834 ft/s 2 

x A  x B  3200 ft

1 1 2 2 (0.76667 ft/s)t AB  (58.667 ft/s)t AB  (0.83447 ft/s 2 )t AB  3200 ft 2 2 2 0.03390t AB  154t AB  3200  0

t  20.685 s and t  4563 s

We have

vB  (vB )0  a B t

At t  t AB :

vB  58.667 ft/s  (0.83447 ft/s 2 )(20.685 s)  75.927 ft/s

t 0

t AB  20.7 s 

v B  51.8 mi/h 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.44 An elevator is moving upward at a constant speed of 4 m/s. A man standing 10 m above the top of the elevator throws a ball upward with a speed of 3 m/s. Determine (a) when the ball will hit the elevator, (b) where the ball will hit the elevator with respect to the location of the man.

SOLUTION Place the origin of the position coordinate at the level of the standing man, the positive direction being up. The ball undergoes uniformly accelerated motion. y B  ( y B ) 0  ( vB ) 0 t 

1 2 gt 2

with ( yB )0  0, (vB )0  3 m/s, and g  9.81 m/s 2 .

yB  3t  4.905t 2 The elevator undergoes uniform motion. y E  ( y E )0  vE t

with ( y E ) 0  10 m and v E  4 m/s. (a)

Set y B  y E

Time of impact.

3t  4.905t 2  10  4t 4.905t 2  t  10  0 t  1.330 s 

t  1.3295 and 1.5334

(b)

Location of impact. y B  (3)(1.3295)  (4.905)(1.3295) 2  4.68 m y E  10  (4)(1.3295)  4.68 m

(checks)

4.68 m below the man 

PROBLEM 11.45 Two rockets are launched at a fireworks display. Rocket A is launched with an initial velocity v0  100 m/s and rocket B is launched t1 seconds later with the same initial velocity. The two rockets are timed to explode simultaneously at a height of 300 m as A is falling and B is rising. Assuming a constant acceleration g  9.81 m/s2 , determine (a) the time t1, (b) the velocity of B relative to A at the time of the explosion.

SOLUTION Place origin at ground level. The motion of rockets A and B is Rocket A:

v A  ( v A ) 0  gt  100  9.81t

y A  ( y A ) 0  (v A ) 0 t  Rocket B:

(1)

1 2 gt  100t  4.905t 2 2

(2)

v B  ( v B ) 0  g (t  t1 )  100  9.81(t  t1 )

(3)

1 g (t  t1 ) 2 2  100(t  t1 )  4.905(t  t1 ) 2

y B  ( y B )0  (vB )0 (t  t1 ) 

(4)

Time of explosion of rockets A and B. y A  y B  300 ft From (2),

300  100t  4.905t 2 4.905t 2  100t  300  0

t  16.732 s and 3.655 s From (4),

300  100(t  t1 )  4.905(t  t12 ) t  t1  16.732 s

Since rocket A is falling, Since rocket B is rising, (a)

Time t1:

(b)

Relative velocity at explosion.

and

3.655 s

t  16.732 s t  t1  3.655 s

t1  t  (t  t1 )

From (1),

v A  100  (9.81)(16.732)  64.15 m/s

From (3),

v B  100  (9.81)(16.732  13.08)  64.15 m/s

Relative velocity:

vB/A  vB  v A

t1  13.08 s 

vB/A  128.3 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.46 Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound lane at a constant speed of 60 mi/h. At t  0, A starts and accelerates at a constant rate a A , while at t  5 s, B begins to slow down with a constant deceleration of magnitude a A /6. Knowing that when the cars pass each other x  294 ft and v A  v B , determine (a) the acceleration a A , (b) when the vehicles pass each other, (c) the distance d between the vehicles at t  0.

SOLUTION

For t  0:

v A  0  a At

1 x A  0  0  a At 2 2

0  t  5 s:

xB  0  (vB )0 t

At t  5 s:

x B  (88 ft/s)(5 s)  440 ft

For t  5 s:

vB  (vB )0  aB (t  5)

1 aB   a A 6

xB  ( xB ) S  (vB )0 (t  5) 

1 aB (t  5) 2 2

( v B ) 0  60 mi/h  88 ft/s

Assume t  5 s when the cars pass each other. At that time (t AB ), v A  vB :

a At AB  (88 ft/s) 

x A  294 ft :

294 ft 

Then or

a A 76 t AB  56  1 2

2 a At AB



1 2 a At AB 2

88 294

2 44t AB  343t AB  245  0

aA (t AB  5) 6

PROBLEM 11.46 (Continued)

Solving (a)

With t AB  5 s,

t AB  0.795 s

294 ft 

and

t AB  7.00 s

1 a A (7.00 s)2 2

a A  12.00 ft/s 2 

or (b)

t AB  7.00 s 

From above

Note: An acceptable solution cannot be found if it is assumed that t AB  5 s. (c)

We have

d  x  ( xB )t AB  294 ft  440 ft  (88 ft/s)(2.00 s) 1 1      12.00 ft/s 2  (2.00 s)2 2 6 

or

d  906 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.47 The elevator shown in the figure moves downward with a constant velocity of 4 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the counterweight W, (c) the relative velocity of the cable C with respect to the elevator, (d ) the relative velocity of the counterweight W with respect to the elevator.

SOLUTION Choose the positive direction downward. (a)

Velocity of cable C. yC  2 y E  constant vC  2 v E  0

v E  4 m/s

But, or (b)

vC  2 v E  8 m/s

v C  8.00 m/s 

Velocity of counterweight W. yW  y E  constant vW  v E  0

(c)

vW   v E  4 m/s

vW  4.00 m/s 

Relative velocity of C with respect to E.

vC/E  vC  vE  (8 m/s)  (4 m/s)  12 m/s

vC/E  12.00 m/s  (d )

Relative velocity of W with respect to E.

vW /E  vW  vE  (4 m/s)  (4 m/s)  8 m/s

vW /E  8.00 m/s 

PROBLEM 11.48 The elevator shown starts from rest and moves upward with a constant acceleration. If the counterweight W moves through 30 ft in 5 s, determine (a) the acceleration of the elevator and the cable C, (b) the velocity of the elevator after 5 s.

SOLUTION We choose positive direction downward for motion of counterweight. yW  At t  5 s,

1 aW t 2 2

yW  30 ft

30 ft 

1 aW (5 s) 2 2

aW  2.4 ft/s 2 (a)

Accelerations of E and C. Since Thus: Also, Thus:

(b)

aW  2.4 ft/s 2

yW  y E  constant

vW  v E  0, and

aW  a E  0

aE  aW  (2.4 ft/s 2 ), yC  2 y E  constant, vC  2 vE  0, and

a E  2.40 ft/s 2  aC  2 a E  0

aC  2aE  2(2.4 ft/s2 )  4.8 ft/s 2 ,

aC  4.80 ft/s 2 

Velocity of elevator after 5 s.

vE  (vE )0  aE t  0  (2.4 ft/s 2 )(5 s)  12 ft/s

( v E ) 5  12.00 ft/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

Problem 11.49 An athlete pulls handle A to the left with a constant velocity of 0.5 m/s. Determine (a) the velocity of the weight B, (b) the relative velocity of weight B with respect to the handle A.

SOLUTION x A  4 y B  constant

From the diagram:

(a)

v A  4vB  0

Differentiate

Sketch:

(1)

x A  0.5 m/s

v B  0.125 m/s

Solve (1) for vB

vB  0.125 m/s  

(b) Finding the relative velocity:

 vB  0.125  m/s  v A  0.5  m/s

   vB / A  vB  v A

 0.125    0.5   m/s

or

v B / A  0.5  0.125  m/s

v B / A  0.5154 m/s

14 

Problem 11.50 An athlete pulls handle A to the left with a constant acceleration. Knowing that after the weight B has been lifted 4 in. its velocity is 2 ft/s, determine (a) the accelerations of handle A and weight B (b) the velocity and change in position of handle A after 0.5 sec.

SOLUTION (a)

From the diagram:

Sketch: x A  3 y B  constant

v A  3v B  0

(1)

a A  3a B  0

(2)

Uniform Acceleration of weight B



v B 2   v B o +2aB y B   y B o 2

Substitute in known values



 2 ft/s 2 =02  2aB  

4  ft  0  12  

aB  6 ft/s2 Using (2)

a A  3(6 ft/s 2 )  0

aA  18 ft/s2  

aB  6 ft/s2   (b) Uniform Acceleration of Handle A

vA   vA o  aAt





v A  0  18 ft/s2  0.5 s  v A  9 ft/s  

or Also

1 y A   y A o   v A o t  a At 2 2

y A   y A o  0  or





1 2 18 ft/s2  0.5 s  2 y A  2.25 ft  

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.51 Slider block B moves to the right with a constant velocity of 300 mm/s. Determine (a) the velocity of slider block A, (b) the velocity of portion C of the cable, (c) the velocity of portion D of the cable, (d ) the relative velocity of portion C of the cable with respect to slider block A.

SOLUTION

x B  ( x B  x A )  2 x A  constant

From the diagram Then

2 v B  3v A  0

(1)

and

2 a B  3a A  0

(2)

 x D  x A  constant

Also, we have

vD  v A  0

Then (a)

Substituting into Eq. (1)

2(300 mm/s)  3v A  0

or (b)

From the diagram Then Substituting

From the diagram Then Substituting or

v A  200 mm/s



v C  600 mm/s



v D  200 mm/s



x B  ( x B  xC )  constant 2 v B  vC  0

2(300 mm/s)  vC  0

or (c)

(3)

( xC  x A )  ( x D  x A )  constant vC  2 v A  v D  0 600 mm/s  2(200 mm/s)  v D  0

PROBLEM 11.51 (Continued)

(d)

We have

vC/A  vC  v A  600 mm/s  200 mm/s

or

vC/A  400 mm/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 11.52 At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slider block A has moved 240 mm to the right its velocity is 60 mm/s, determine (a) the accelerations of A and B, (b) the acceleration of portion D of the cable, (c) the velocity and change in position of slider block B after 4 s.

SOLUTION

x B  ( x B  x A )  2 x A  constant

From the diagram Then

2v B  3v A  0

(1)

and

2 a B  3a A  0

(2)

(a)

First observe that if block A moves to the right, v A  and Eq. (1)  v B  . Then, using Eq. (1) at t  0 2(150 mm/s)  3(v A ) 0  0 (v A ) 0  100 mm/s

or

Also, Eq. (2) and a B  constant  a A  constant

vA2  (vA )02  2a A [ xA  ( xA )0 ]

Then When x A  ( x A ) 0  240 mm:

(60 mm/s)2  (100 mm/s)2  2a A (240 mm) or or

aA  

40 mm/s 2 3

a A  13.33 mm/s 2



PROBLEM 11.52 (Continued)

Then, substituting into Eq. (2)  40  2 aB  3   mm/s 2   0  3 

aB  20 mm/s 2

or (b)

a B  20.0 mm/s2



From the diagram,  xD  x A  constant vD  v A  0

Then Substituting

aD  a A  0

 40  mm/s 2   0 aD     3 

or (c)

We have

v B  (v B ) 0  a B t

At t  4 s:

vB  150 mm/s  (20.0 mm/s2 )(4 s)

or Also At t  4 s:

xB  ( xB ) 0  ( vB ) 0 t 



v B  70.0 mm/s



1 aB t 2 2

xB  ( xB )0  (150 mm/s)(4 s) 

or

a D  13.33 mm/s 2

1 (20.0 mm/s 2 )(4 s) 2 2 x B  ( x B ) 0  440 mm

Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 11.53 A farmer wants to get his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Determine the velocity and acceleration of the hay bale when the horse is 10 ft away from the barn.

SOLUTION From the diagram, we can write an expression for the total length of rope:

L  20  yB  xH2  (20)2 Differentiate

dL 0 dt

xH x H

0   y B  Velocities

vB  y B and vH  xH vB 

Differentiate again:

x  (20) 2 2 H

aB 

aB 

or

aB 

at xH  10

vB 

xH v H xH2  (20) 2 dvB dt

x H vH xH2  (20) 2

vH2 xH2  (20) 2

xH vH



xH2  (20) 2

xH a H



xH2  (20) 2

aB 



xH vH (  12 )(2 xH ) xH

x

x

2 H

 (20) 2 

3/ 2

xH2 vH2

2 H

 (20) 2 

3/2

10(1)

10    20  2

vB  0.447 ft/s 

2

1 2 2 10    20  2

and





10  0  2 2 10    20 

aB  0.0447  0  0.000894

10  1  3/2 102   202  2

2

aB  0.0358 ft/s2 

PROBLEM 11.54 The motor M reels in the cable at a constant rate of 100 mm/s. Determine (a) the velocity of load L, (b) the velocity of pulley B with respect to load L.

SOLUTION

Let xB and xL be the positions, respectively, of pulley B and load L measured downward from a fixed elevation above both. Let xM be the position of a point on the cable about to enter the reel driven by the motor. Then, considering the lengths of the two cables,

xM  3 xB  constant

vM  3vB  0

xL  ( xL  xB )  constant

2vL  vB  0

vM  100 mm/s

with

vB   vL 

vM  33.333 m/s 3

vB  16.667 mm/s 2 v L  16.67 mm/s 

(a)

Velocity of load L.

(b)

Velocity of pulley B with respect to load L. vB/L  vB  vL  33.333  (16.667)  16.667

v B/L  16.67 mm/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.55 Collar A starts from rest at t = 0 and moves upward with a constant acceleration of 3.6 in./s2. Knowing that collar B moves downward with a constant velocity of 18 in./s, determine (a) the time at which the velocity of block C is zero, (b) the corresponding position of block C.

SOLUTION Define positions as positive downward from a fixed level. Constraint of cable.

 yB  y A    yC

 y A   2  yC  yB   constant

3 yC  y B  2 y A  constant

Differentiate

3vC  vB  2v A  0

(1)

Differentiate again

3aC  a B  2 a A  0

(2)

Motion of block C:

 vA 0  0,

aA   3.6 in./s2, vB   vB 0  18 in./s, aB  0

1  vB   2  v A    6 in./s 0 0 3

Using (1)

 vC 0 

Using (2)

aC 

Constant acceleration

vC   vC 0  aC t

Constant acceleration

xC   xC 0   vC 0 t 

1 1  aB  2a A   0   2   3.6    2.4 in./s2 3 3

(3)

 6  1.2t

 6t  0.6t

1 aC t 2 2

(4)

2

(a) Time at vC  0 : Using (3)

0  6  2.4t

t  2.5 s 

(b) Corresponding position of block C: Using (4)

2 1 xC   xC 0   6  2.5      2.4  2.5  2  

xC   xC 0  7.5 in.  

PROBLEM 11.56 Block A starts from rest at t  0 and moves downward with a constant acceleration of 6 in./s2. Knowing that block B moves up with a constant velocity of 3 in./s, determine (a) the time when the velocity of block C is zero, (b) the corresponding position of block C.

SOLUTION

The cable lengths are constant.

L1  2 yC  2 yD  constant L 2  y A  yB  ( yB  yD )  constant Eliminate yD.

L1  2 L 2  2 yC  2 yD  2 y A  2 yB  2( yB  yD )  constant 2( yC  y A  2 yB )  constant Differentiate to obtain relationships for velocities and accelerations, positive downward. vC  v A  2 v B  0

(1)

aC  a A  2 a B  0

(2)

Use units of inches and seconds. Motion of block A:

v A  a A t  6t

y A  Motion of block B:

1 1 a At 2  (6)t 2  3t 2 2 2

v B  3 in./s

v B  3 in./s

 y B  v B t  3t

Motion of block C:

From (1),

vC  v A  2vB  6t  2(3)  6  6t yC  (a)

Time when vC is zero.

(b)

Corresponding position.



t 0

vC dt  6t  3t 2

6  6t  0

yC  (6)(1)  (3)(1)2  3 in.

t  1.000 s 

 yC  3.00 in. 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.57 Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s 2 . Knowing that at t  2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of A and C, (c) the change in position of slider block C after 3 s.

SOLUTION

From the diagram 3 y A  4 y B  xC  constant

Then

3v A  4v B  vC  0

(1)

and

3a A  4 a B  aC  0

(2)

( v B )  0,

Given:

a A  constent (aC )  75 mm/s 2

At t  2 s,

v B  480 mm/s v C  280 mm/s

(a)

Eq. (2) and a A  constant and aC  constant  a B  constant vB  0  aB t

Then At t  2 s:

480 mm/s  a B (2 s)

aB  240 mm/s2 

or

a B  240 mm/s2 

or

a A  345 mm/s2 

Substituting into Eq. (2)

3a A  4(240 mm/s 2 )  (75 mm/s 2 )  0 a A  345 mm/s

  

PROBLEM 11.57 (Continued)

(b)

vC  ( vC ) 0  aC t

We have At t  2 s:

280 mm/s  (vC ) 0  (75 mm/s)(2 s) vC  130 mm/s



or

( v C ) 0  130.0 mm/s



Then, substituting into Eq. (1) at t  0 3( v A ) 0  4(0)  (130 mm/s)  0 v A  43.3 mm/s

(c)

We have At t  3 s:

xC  ( xC )0  (vC )0 t 

or 1 aC t 2 2

xC  ( xC )0  (130 mm/s)(3 s)   728 mm

(v A ) 0  43.3 mm/s 

1 (75 mm/s 2 )(3 s) 2 2 or

x C  ( x C ) 0  728 mm

Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 11.58 Block B moves downward with a constant velocity of 20 mm/s. At t  0, block A is moving upward with a constant acceleration, and its velocity is 30 mm/s. Knowing that at t  3 s slider block C has moved 57 mm to the right, determine (a) the velocity of slider block C at t  0, (b) the accelerations of A and C, (c) the change in position of block A after 5 s.

SOLUTION

From the diagram 3 y A  4 y B  xC  constant

Then

3v A  4v B  vC  0

(1)

and

3a A  4 a B  aC  0

(2)

v B  20 mm/s ;

Given:

( v A ) 0  30 mm/s

(a)

Substituting into Eq. (1) at t  0 3( 30 mm/s)  4(20 mm/s)  ( vC ) 0  0 ( vC ) 0  10 mm/s

(b)

We have At t  3 s:

( v C ) 0  10.00 mm/s

or

xC  ( xC )0  (vC )0 t 

1 aC t 2 2

57 mm  (10 mm/s)(3 s) 

1 aC (3 s) 2 2

aC  6 mm/s2 

Now



v B  constant  a B  0

or

aC  6.00 mm/s 2



PROBLEM 11.58 (Continued)

Then, substituting into Eq. (2)

3a A  4(0)  (6 mm/s 2 )  0 a A  2 mm/s2 (c)

We have At t  5 s:

y A  ( y A ) 0  (v A ) 0 t 

or

a A  2.00 mm/s 2 

1 a At 2 2

y A  ( y A )0  (30 mm/s)(5 s) 

1 (2 mm/s 2 )(5 s)2 2

 175 mm or

y A  ( y A ) 0  175.0 mm 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.59 The system shown starts from rest, and each component moves with a constant acceleration. If the relative acceleration of block C with respect to collar B is 60 mm/s 2 upward and the relative acceleration of block D with respect to block A is 110 mm/s 2 downward, determine (a) the velocity of block C after 3 s, (b) the change in position of block D after 5 s.

SOLUTION From the diagram 2 y A  2 y B  yC  constant

Cable 1: Then

2v A  2vB  vC  0

(1)

and

2 a A  2 a B  aC  0

(2)

( y D  y A )  ( y D  y B )  constant

Cable 2: Then and

(3)

 a A  aB  2aD  0

(4)

At t  0, v  0; all accelerations constant; aC/B  60 mm/s2 , aD /A  110 mm/s 2

Given: (a)

 v A  vB  2vD  0

We have

aC/B  aC  a B  60

or

a B  aC  60

and

a D/A  a D  a A  110

or

a A  a D  110

Substituting into Eqs. (2) and (4) Eq. (2): or Eq. (4): or

2( a D  110)  2( aC  60)  aC  0 3aC  2 a D  100

(5)

 ( a D  110)  ( aC  60)  2 a D  0  aC  a D  50

(6)

PROBLEM 11.59 (Continued)

Solving Eqs. (5) and (6) for aC and a D

aC  40 mm/s2 aD  10 mm/s 2 Now

vC  0  aC t

At t  3 s:

vC  (40 mm/s 2 )(3 s) v C  120.0 mm/s 

or (b)

We have At t  5 s: or

yD  ( yD )0  (0)t  yD  ( yD )0 

1 aD t 2 2

1 (10 mm/s 2 )(5 s)2 2 y D  ( y D ) 0  125.0 mm 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.60* The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine (a) the accelerations of A and B if aB  10 mm/s2 , (b) the change in position of block D when the velocity of block C is 600 mm/s upward.

SOLUTION From the diagram Cable 1:

2 y A  2 y B  yC  constant

Then

2v A  2vB  vC  0

(1)

and

2 a A  2 a B  aC  0

(2)

Cable 2: ( y D  y A )  ( y D  y B )  constant Then

 v A  vB  2vD  0

(3)

and

 a A  aB  2aD  0

(4)

Given: At

t0 v0 ( y A )0  ( yB )0  ( yC )0

All accelerations constant. At t  2 s yC /A  280 mm

When

v B /A  80 mm/s y A  ( y A ) 0  160 mm y B  ( y B ) 0  320 mm

aB  10 mm/s 2

PROBLEM 11.60* (Continued)

(a)

We have

y A  ( y A )0  (0)t 

1 a At 2 2

and

yC  ( yC )0  (0)t 

1 aC t 2 2

Then

yC/A  yC  y A 

1 (aC  a A )t 2 2

At t  2 s, yC/A  280 mm: 280 mm 

or

1 ( aC  a A )(2 s) 2 2

aC  a A  140

(5)

Substituting into Eq. (2) 2 a A  2 a B  ( a A  140)  0

or

1 a A  (140  2aB ) 3

Now

vB  0  a B t

(6)

v A  0  a At v B /A  v B  v A  ( a B  a A ) t

Also When



yB  ( yB )0  (0)t  v B /A  80 mm/s :

1 aB t 2 2 80  ( a B  a A )t

 y A  160 mm :

160 

1 a At 2 2

y B  320 mm :

320 

1 aB t 2 2

(7)

1 (aB  a A )t 2 2

Then

160 

Using Eq. (7)

320  (80)t or t  4 s

Then

160 

1 a A (4)2 2

or

a A  20.0 mm/s2 

and

320 

1 aB (4)2 2

or

a B  40.0 mm/s2 

Note that Eq. (6) is not used; thus, the problem is over-determined.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.60* (Continued)

(b)

Substituting into Eq. (5)

aC  20  140  120 mm/s 2 and into Eq. (4)

(20 mm/s 2 )  (40 mm/s 2 )  2aD  0 or

aD  30 mm/s 2

Now

vC  0  aC t

When vC  600 mm/s: or Also At t  5 s: or

600 mm/s  (120 mm/s 2 )t t 5s yD  ( yD )0  (0)t 

y D  ( y D )0 

1 aD t 2 2

1 (30 mm/s 2 )(5 s) 2 2 y D  ( y D ) 0  375 mm 

PROBLEM 11.61 A particle moves in a straight line with a constant acceleration of 4 ft/s 2 for 6 s, zero acceleration for the next 4 s, and a constant acceleration of 4 ft/s 2 for the next 4 s. Knowing that the particle starts from the origin and that its velocity is 8 ft/s during the zero acceleration time interval, (a) construct the v  t and x  t curves for 0  t  14 s, (b) determine the position and the velocity of the particle and the total distance traveled when t  14 s.

SOLUTION From the a  t curve A1  24 ft/s, A2  16 ft/s

(a) Construct the v  t curve :

v6   8 ft/s

v0  v6  A1  8   24   16 ft/s v10   8 ft/s

v14  v10  A2  8  16



 8 ft/s

From the v  t curve

A3  32 ft, A4  8 ft A5   32 ft, A6  8 ft , A7  8 ft

(b) Construct the x  t curve

x0  0 x4  x0  A3  32 ft

x6  x4  A4  24 ft x10  x6  A5  8 ft x12  x10  A6   16 ft



x14  x12  A7   8 ft

Total Distance Traveled

0  t  4 s,

d1  32  0  32 ft

4 s  t  12 s,

d 2  16  32  48 ft

12 s  t  14 s,

d3  8   16  8 ft

d  32  48  8

d  88 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

Problem 11.62 A particle moves in a straight line with a constant acceleration of 4 ft/s 2 for 6 s, zero acceleration for the next 4 s, and a constant acceleration of 4 ft/s 2 for the next 4 s. Knowing that the particle starts from the origin with v0  16 ft/s, (a) construct the v  t and x  t curves for 0  t  14 s, (b) determine the amount of time during which the particle is further than 16 ft from the origin.

SOLUTION From the a  t curve A1  24 ft/s, A2  16 ft/s

(b) Construct the v  t curve :

v0  16 ft/s

v6  v0  A1  16   24   8 ft/s v10  v6   8 ft/s

v14  v10  A2  8  16



 8 ft/s

From the v  t curve

A3  32 ft, A4  8 ft

A5   32 ft, A6  8 ft , A7  8 ft

Construct the x  t curve

x0  0 x4  x0  A3  32 ft x6  x4  A4  24 ft x10  x6  A5  8 ft

x12  x10  A6   16 ft



x14  x12  A7   8 ft

(b)

Time for x  16 ft.

From the x  t diagram, this is time interval t1 to t 2 . From 0  t  6 s,

dx  v  16  4t dt

Integrating, using limits x  0 when t  0 and x  16 ft when t  t1

 x160

t

 16t  2t 2  0

or

16  16t1  2t12

PROBLEM 11.62 (Continued) t12  8t1  8  0

or Using the quadratic formula:

t1 

8

82   4 18   2 1

 4  2.828  1.172 s

and

6.828 s

The larger root is out of range, thus t1  1.172 s From 6  t  10,

x  24  8  t  6   72  8t

Setting x  16,

16  72  8t 2

Time Interval:

t  t2  t1

or

t2  7 s

t  5.83 s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.63 A particle moves in a straight line with the velocity shown in the figure. Knowing that x  540 m at t  0, (a) construct the a t and x t curves for 0  t  50 s, and determine (b) the total distance traveled by the particle when t  50 s, (c) the two times at which x  0.

SOLUTION (a)

at  slope of v  t curve at time t

From t  0 to t  10 s:

v  constant  a  0 20  60  5 m/s 2 26  10

t  10 s to t  26 s:

a

t  26 s to t  41 s:

v  constant  a  0

t  41 s to t  46 s:

a

t  46 s:

5  (20)  3 m/s 2 46  41

v  constant  a  0

x2  x1  (area under v t curve from t1 to t 2 )

At t  10 s:

x10  540  10(60)  60 m

Next, find time at which v  0. Using similar triangles tv  0  10 60 At

t  22 s: t  26 s: t  41 s: t  46 s: t  50 s:



26  10 80

1 x22  60  (12)(60)  420 m 2 1 x26  420  (4)(20)  380 m 2 x41  380  15(20)  80 m  20  5  x46  80  5    17.5 m  2  x50  17.5  4(5)  2.5 m

or

tv  0  22 s

PROBLEM 11.63 (Continued)

(b)

From t  0 to t  22 s: Distance traveled  420  (540)  960 m  t  22 s to t  50 s: Distance traveled  | 2.5  420|  422.5 m Total distance traveled  (960  422.5) ft  1382.5 m Total distance traveled  1383 m 

(c)

Using similar triangles Between 0 and 10 s:

(t x 0 )1  0 10  540 600 (t x  0 )1  9.00 s 

Between 46 s and 50 s:

(t x  0 ) 2  46 4  17.5 20 (t x  0 ) 2  49.5 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.64 A particle moves in a straight line with the velocity shown in the figure. Knowing that x  540 m at t  0, (a) construct the a  t and x  t curves for 0  t  50 s, and determine (b) the maximum value of the position coordinate of the particle, (c) the values of t for which the particle is at x  100 m.

SOLUTION (a)

at  slope of v  t curve at time t

From t  0 to t  10 s:

v  constant  a  0

20  60  5 m/s 2 26  10

t  10 s to t  26 s:

a

t  26 s to t  41 s:

v  constant  a  0

t  41 s to t  46 s:

a

t  46 s:

5  (20)  3 m/s2 46  41

v  constant  a  0

x2  x1  (area under v t curve from t1 to t 2 )

At t  10 s:

x10  540  10(60)  60 m

Next, find time at which v  0. Using similar triangles tv  0  10 60 At

t  22 s: t  26 s: t  41 s: t  46 s: t  50 s:



26  10 80

or

1 x22  60  (12)(60)  420 m 2 1 x26  420  (4)(20)  380 m 2 x41  380  15(20)  80 m  20  5  x46  80  5    17.5 m  2  x50  17.5  4(5)  2.5 m

tv  0  22 s

PROBLEM 11.64 (Continued)

(b)

Reading from the x  t curve

(c)

Between 10 s and 22 s

xmax  420 m 

100 m  420 m  (area under v t curve from t , to 22 s) m 1 100  420  (22  t1 )(v1 ) 2 (22  t1 )( v1 )  640

Using similar triangles v1 60  22  t1 12 Then

v1  5(22  t1 )

or

(22  t1 )[5(22  t1 )]  640

t1  10.69 s

and

t1  33.3 s

10 s  t1  22 s 

We have

t1  10.69 s 

Between 26 s and 41 s: Using similar triangles 41  t2 15  20 300 t 2  40.0 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

Problem 11.65 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = – 48 ft at t = 0, draw the a–t and x–t curves for 0 < t < 40 s and determine (a) the maximum value of the position coordinate of the particle, (b) the values of t for which the particle is at a distance of 108 ft from the origin.

SOLUTION The a  t curve is the slope of the v  t curve

0  t  10 s,

a0

10 s < t  18 s,

a

18  6  1.5 ft/s 2 18  10

18 s < t  30 s,

a

18  18   3 ft/s 2 30  18

30 s < t  40 s

a0



Points on the x–t curve may be calculated using areas of the v–t curve. A1  (10)(6)  60 ft

A2 

1 (6  18)(18  10)  96 ft 2

A3 

1 (18)(24  18)  54 ft 2

A4 

1 (18)(30  24)   54 ft 2

A5  ( 18)(40  30)   180 ft

Value of x at specific times:

x0   48 ft

x10  x0  A1  12 ft x18  x10  A2  108 ft x24  x18  A3  162 ft x30  x24  A4  108 ft x40  x30  A5   72 ft



PROBLEM 11.65 (Continued) (a) Maximum value of x occurs when: v  0, i.e. t  24 s.

xmax  162 ft 

(b) Times when x  108 ft. From the x  t curve

t  18 s and t  30 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.66 A parachutist is in free fall at a rate of 200 km/h when he opens his parachute at an altitude of 600 m. Following a rapid and constant deceleration, he then descends at a constant rate of 50 km/h from 586 m to 30 m, where he maneuvers the parachute into the wind to further slow his descent. Knowing that the parachutist lands with a negligible downward velocity, determine (a) the time required for the parachutist to land after opening his parachute, (b) the initial deceleration.

SOLUTION Assume second deceleration is constant. Also, note that 200 km/h  55.555 m/s, 50 km/h  13.888 m/s

PROBLEM 11.66 (Continued)

(a)

Now  x  area under v t curve for given time interval Then

 55.555  13.888  (586  600) m  t1   m/s 2  

t1  0.4032 s (30  586) m  t2 (13.888 m/s) t2  40.0346 s 1 (0  30) m   (t3 )(13.888 m/s) 2 t3  4.3203 s t total  (0.4032  40.0346  4.3203) s

(b)

We have

ainitial  

t total  44.8 s 

 vinitial t1 [13.888  (55.555)] m/s 0.4032 s

 103.3 m/s2

ainitial  103.3 m/s 2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.67 A commuter train traveling at 40 mi/h is 3 mi from a station. The train then decelerates so that its speed is 20 mi/h when it is 0.5 mi from the station. Knowing that the train arrives at the station 7.5 min after beginning to decelerate and assuming constant decelerations, determine (a) the time required for the train to travel the first 2.5 mi, (b) the speed of the train as it arrives at the station, (c) the final constant deceleration of the train.

SOLUTION Given: At t  0, v  40 mi/h, x  0; when x  2.5 mi, v  20 mi/h; at t  7.5 min, x  3 mi; constant decelerations. The v  t curve is first drawn as shown. (a)

A1  2.5 mi

We have

1h  40  20  (t1 min)   2.5 mi  mi/h  60 min  2  t1  5.00 min 

(b)

A2  0.5 mi

We have

 20  v2 (7.5  5) min    2

1h   mi/h  60 min  0.5 mi  v2  4.00 mi/h 

(c)

We have

afinal  a12 

(4  20) mi/h 5280 ft 1 min 1h    (7.5  5) min mi 60 s 3600 s

afinal  0.1564 ft/s2 

PROBLEM 11.68 A temperature sensor is attached to slider AB which moves back and forth through 60 in. The maximum velocities of the slider are 12 in./s to the right and 30 in./s to the left. When the slider is moving to the right, it accelerates and decelerates at a constant rate of 6 in./s2; when moving to the left, the slider accelerates and decelerates at a constant rate of 20 in./s2. Determine the time required for the slider to complete a full cycle, and construct the v – t and x  t curves of its motion.

SOLUTION The v  t curve is first drawn as shown. Then ta 

vright aright



12 in./s 2s 6 in./s 2

vleft 30 in./s  aleft 20 in./s  1.5 s

td 

A1  60 in.

Now

[(t1  2) s](12 in./s)  60 in.

or

t1  7 s

or

A2  60 in.

and or

{[(t 2  7)  1.5] s}(30 in./s)  60 in.

or

t 2  10.5 s

tcycle  t2

Now

tcycle  10.5 s 

We have xii  xi  (area under v  t curve from ti to tii ) At

t  5 s:

1 (2) (12)  12 in. 2 x5  12  (5  2)(12)

t  7 s:

x7  60 in.

t  2 s:

x2 

 48 in.

t  8.5 s: t  9 s:

1 x8.5  60  (1.5)(30) 2  37.5 in. x9  37.5  (0.5)(30)

 22.5 in. t  10.5 s:

x10.5  0

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.69 In a water-tank test involving the launching of a small model boat, the model’s initial horizontal velocity is 6 m/s, and its horizontal acceleration varies linearly from 12 m/s 2 at t  0 to 2 m/s 2 at t  t1 and then remains equal to 2 m/s 2 until t  1.4 s. Knowing that v  1.8 m/s when t  t1 , determine (a) the value of t1 , (b) the velocity and the position of the model at t  1.4 s.

SOLUTION Given:

v0  6 m/s; for 0  t  t1 ,

for

t1  t  1.4 s a  2 m/s2 ;

at

t  0 a  12 m/s 2 ;

at t  t1

a  2 m/s 2 , v  1.8 m/s 2

The a t and vt curves are first drawn as shown. The time axis is not drawn to scale. (a)

We have

vt1  v0  A1  12  2  2 1.8 m/s  6 m/s  (t1 s)   m/s  2  t1  0.6 s 

(b)

We have

v1.4  vt1  A2 v1.4  1.8 m/s  (1.4  0.6) s  2 m/s 2 v1.4  0.20 m/s 

Now x1.4  A3  A4 , where A3 is most easily determined using integration. Thus, for 0  t  t1 : Now

a

2  (12) 50 t  12  t  12 0.6 3

dv 50  a  t  12 dt 3

PROBLEM 11.69 (Continued)

At t  0, v  6 m/s:



v 6

dv 



t

50  t  12  dt  0 3 

v6

or

25 2 t  12t 3

We have

dx 25  v  6  12t  t 2 dt 3

Then

A3 



xt1 0

dx 



0.6 0

(6  12t 

25 2 t )dt 3

0.6

25    6t  6t 2  t 3   2.04 m 9 0  Also Then or

 1.8  0.2  A4  (1.4  0.6)    0.8 m 2   x1.4  (2.04  0.8) m x1.4  2.84 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.70 The acceleration record shown was obtained for a small airplane traveling along a straight course. Knowing that x  0 and v  60 m/s when t  0, determine (a) the velocity and position of the plane at t  20 s, (b) its average velocity during the interval 6 s  t  14 s.

SOLUTION

Geometry of “bell-shaped” portion of vt curve

The parabolic spandrels marked by * are of equal area. Thus, total area of shaded portion of vt diagram is: = x = 6 m

(a)

When t  20 s:

v20  60 m/s  x20  (60 m/s) (20 s)  (shaded area)

 1200 m  6 m (b)

From t  6 s to t  14 s:

x20  1194 m 

t  8 s  x  (60 m/s) (14 s  6 s)  (shaded area)  (60 m/s)(8 s)  6 m  480 m  6 m  474 m vaverage 

 x 474 m  8s t

vaverage  59.25 m/s 

PROBLEM 11.71 In a 400-m race, runner A reaches her maximum velocity v A in 4 s with constant acceleration and maintains that velocity until she reaches the half-way point with a split time of 25 s. Runner B reaches her maximum velocity v B in 5 s with constant acceleration and maintains that velocity until she reaches the half-way point with a split time of 25.2 s. Both runners then run the second half of the race with the same constant deceleration of 0.1 m/s 2 . Determine (a) the race times for both runners, (b) the position of the winner relative to the loser when the winner reaches the finish line.

SOLUTION Sketch v t curves for first 200 m. Runner A:

t1  4 s, t 2  25  4  21 s

A1 

1 (4)(v A )max  2(v A )max 2

A2  21(v A ) max A1  A2   x  200 m 23(v A ) max  200

Runner B:

or

t1  5 s,

A1 

(v A ) max  8.6957 m/s

t 2  25.2  5  20.2 s

1 (5)(vB ) max  2.5(vB ) max 2

A2  20.2(v B ) max

A1  A2   x  200 m 22.7(vB ) max  200

Sketch v t curve for second 200 m. A3  vmaxt3 

t3  Runner A:

or

(v B ) max  8.8106 m/s

v  | a |t3  0.1t3

1 vt3  200 2

or

vmax  (vmax )2  (4)(0.05)(200)

(vmax ) A  8.6957,

Reject the larger root. Then total time

(2)(0.05) (t3 ) A  146.64 s

0.05t32  vmaxt3  200  0



 10 vmax  (vmax )2  40 and



27.279 s

t A  25  27.279  52.279 s

t A  52.2 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.71 (Continued)

Runner B: (vmax ) B  8.8106, (t3 ) B  149.45 s Reject the larger root. Then total time

and

26.765 s t B  25.2  26.765  51.965 s t B  52.0 s 

Velocity of A at t  51.965 s: v1  8.6957  (0.1)(51.965  25)  5.999 m/s

Velocity of A at t  51.279 s: v2  8.6957  (0.1)(52.279  25)  5.968 m/s

Over 51.965 s  t  52.965 s, runner A covers a distance  x  x  vave (t ) 

1 (5.999  5.968)(52.279  51.965) 2

 x  1.879 m 

PROBLEM 11.72 A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s 2 and the maximum deceleration obtained by applying the brakes is 20 ft/s 2 . What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h? Draw the vt curve.

SOLUTION Relative to truck, car must move a distance: Allowable increase in speed:

 x  16  40  50  40  146 ft  vm  50  35  15 mi/h  22 ft/s

Acceleration Phase:

t1  22/5  4.4 s

A1 

1 (22)(4.4)  48.4 ft 2

Deceleration Phase:

t3  22/20  1.1 s

A3 

1 (22)(1.1)  12.1 ft 2

But:  x  A1  A2  A3 :

146 ft  48.4  (22)t 2  12.1 t total  t1  t 2  t3  4.4 s  3.89 s  1.1 s  9.39 s

t 2  3.89 s t B  9.39 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.73 Solve Problem 11.72, assuming that the driver of the car does not pay any attention to the speed limit while passing and concentrates on reaching position B and resuming a speed of 35 mi/h in the shortest possible time. What is the maximum speed reached? Draw the vt curve.

PROBLEM 11.72 A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s 2 and the maximum deceleration obtained by applying the brakes is 20 ft/s 2 . What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h? Draw the vt curve.

SOLUTION Relative to truck, car must move a distance:

 x  16  40  50  40  146 ft

vm  5t1  20t2 ;  x  A1  A2 :

t2 

146 ft 

1 (vm )(t1  t2 ) 2

146 ft 

1 1   (5t1 )  t1  t1  2 4  

t12  46.72

1 t1 4

t1  6.835 s

t2 

1 t1  1.709 4

t total  t1  t 2  6.835  1.709

t B  8.54 s 

vm  5t1  5(6.835)  34.18 ft/s  23.3 mi/h Speed vtotal  35 mi/h, vm  35 mi/h  23.3 mi/h

vm  58.3 mi/h 

PROBLEM 11.74 Car A is traveling on a highway at a constant speed (v A ) 0  60 mi/h, and is 380 ft from the entrance of an access ramp when car B enters the acceleration lane at that point at a speed (v B ) 0  15 mi/h. Car B accelerates uniformly and enters the main traffic lane after traveling 200 ft in 5 s. It then continues to accelerate at the same rate until it reaches a speed of 60 mi/h, which it then maintains. Determine the final distance between the two cars.

SOLUTION (v A )0  60 mi/h, (vB )0  1.5 mi/h; at t  0,

Given:

( x A )0  380 ft, ( xB )0  0; at t  5 s, xB  200 ft; for 15 mi/h  vB  60 mi/h, aB  constant; for vB  60 mi/h,

First note

aB  0 60 mi/h  88 ft/s

15 mi/h  22 ft/s The vt curves of the two cars are then drawn as shown. Using the coordinate system shown, we have at t  5 s, xB  200 ft :

 22 + (vB )5  (5 s)   ft/s  200 ft 2   ( v B ) 5  58 ft/s

or Then, using similar triangles, we have

(88  22) ft/s (58  22) ft/s (  aB )  5s t1 t1  9.16 67 s

or Finally, at t  t1

   22 + 88  ft/s  xB/A  xB  x A  (9.1667 s)    2     [380 ft  (9.1667 s) (88 ft/s)]

or

xB/A  77.5 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.75 An elevator starts from rest and moves upward, accelerating at a rate of 1.2 m/s2 until it reaches a speed of 7.8 m/s, which it then maintains. Two seconds after the elevator begins to move, a man standing 12 m above the initial position of the top of the elevator throws a ball upward with an initial velocity of 20 m/s. Determine when the ball will hit the elevator.

SOLUTION Given:

At t  0 vE  0; For 0  vE  7.8 m/s, aE  1.2 m/s 2 ; For v E  7.8 m/s, a E  0; At t  2 s, v B  20m/s 

The vt curves of the ball and the elevator are first drawn as shown. Note that the initial slope of the curve for the elevator is 1.2 m/s 2 , while the slope of the curve for the ball is  g (9.81 m/s 2 ).

The time t1 is the time when v E reaches 7.8 m/s. Thus, or or

vE  (0)  a E t

7.8 m/s  (1.2 m/s 2 )t1 t1  6.5 s

The time ttop is the time at which the ball reaches the top of its trajectory. Thus, or or

vB  (vB ) 0  g (t  2)

0  20 m/s  (9.81 m/s2 )(ttop  2) s ttop  4.0387 s

PROBLEM 11.75 (Continued)

Using the coordinate system shown, we have 0  t  t1 :

1  yE  12 m   aE t 2  m 2 

At t  ttop :

yB 

and

yE  12 m 

1 (4.0387  2) s  (20 m/s) 2  20.387 m 1 (1.2 m/s 2 )(4.0387 s) 2 2  2.213 m

t  [2  2(4.0387  2)] s  6.0774 s, y B  0

At and at t  t1 ,

yE  12 m 

1 (6.5 s) (7.8 m/s)  13.35 m 2

The ball hits the elevator ( y B  y E ) when ttop  t  t1. For t  ttop :

1  yB  20.387 m   g (t  ttop )2  m 2  

Then, yB  yE

when

1 20.387 m  (9.81 m/s 2 ) (t  4.0387) 2 2 1  12 m  (1.2 m/s 2 ) (t s)2 2 5.505t 2  39.6196t  47.619  0

or Solving Since 1.525 s is less than 2 s,

t  1.525 s and t  5.67 s

t  5.67 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.76 Car A is traveling at 40 mi/h when it enters a 30 mi/h speed zone. The driver of car A decelerates at a rate of 16 ft/s2 until reaching a speed of 30 mi/h, which she then maintains. When car B, which was initially 60 ft behind car A and traveling at a constant speed of 45 mi/h, enters the speed zone, its driver decelerates at a rate of 20 ft/s2 until reaching a speed of 28 mi/h. Knowing that the driver of car B maintains a speed of 28 mi/h, determine (a) the closest that car B comes to car A, (b) the time at which car A is 70 ft in front of car B.

SOLUTION (v A )0  40 mi/h; For 30 mi/h  v A  40 mi/h, a A  16 ft/s 2 ; For v A  30 mi/h, a A  0;

Given:

( x A /B )0  60 ft; (vB )0  45 mi/h; When xB  0, aB  20 ft/s 2 ; For vB  28 mi/h, aB  0 First note

40 mi/h  58.667 ft/s 30 mi/h  44 ft/s 45 mi/h  66 ft/s 28 mi/h  41.067 ft/s

At t  0

The vt curves of the two cars are as shown. At

t  0:

Car A enters the speed zone.

t  (t B )1:

Car B enters the speed zone.

t  tA:

Car A reaches its final speed.

t  tmin :

vA  vB

t  (tB )2 :

Car B reaches its final speed.

PROBLEM 11.76 (Continued)

(a)

(v A )final  (v A )0 tA

aA 

We have

16 ft/s 2 

or

(44  58.667) ft/s tA

t A  0.91669 s

or Also

60 ft  (t B )1 (vB )0

or

60 ft  (t B )1 (66 ft/s) aB 

and

20 ft/s 2 

or

or

(t B )1  0.90909 s

(vB )final  (vB )0 (t B ) 2  (t B )1 (41.067  66) ft/s [(tB )2  0.90909] s

Car B will continue to overtake car A while vB  v A . Therefore, ( x A/B )min will occur when v A  vB , which occurs for (tB )1  tmin  (t B ) 2 For this time interval

v A  44 ft/s vB  (vB )0  aB [t  (t B )1 ] Then

at t  tmin :

44 ft/s  66 ft/s  (20 ft/s2 )(tmin  0.90909) s tmin  2.00909 s

or

Finally ( x A/B ) min  ( x A )t  ( xB )t min min   (v )  (v A )final    t A  A 0  (tmin  t A )(v A )final   2       (v )  (v A )final    ( xB )0  (t B )1 (vB )0  [tmin  (t B )1 ]  B 0  2   

   58.667  44   (0.91669 s)  ft/s  (2.00909  0.91669) s  (44 ft/s)   2       66  44     60 ft  (0.90909 s)(66 ft/s)  (2.00909  0.90909) s    ft/s   2     (47.057  48.066) ft  (60  60.000  60.500) ft

 34.623 ft

or ( x A/B )min  34.6 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.76 (Continued)

(b)

Since ( xA/B )  60 ft for t  tmin , it follows that x A/B  70 ft for t  (tB )2 [Note (t B ) 2

tmin ]. Then, for t  (tB )2

x A/B  ( xA/B )min  [(t  tmin )(v A )final ]    (vB )final   (v )  [(tB )2  (tmin )]  A final  [t  (tB )2 ](vB )final   2     or

70 ft  34.623 ft  [(t  2.00909) s  (44 ft/s)]    44  41.06   (2.15574  2.00909) s   ft/s  (t  2.15574) s  (41.067) ft/s   2    

or

t  14.14 s 

PROBLEM 11.77 An accelerometer record for the motion of a given part of a mechanism is approximated by an arc of a parabola for 0.2 s and a straight line for the next 0.2 s as shown in the figure. Knowing that v  0 when t  0 and x  0.8 ft when t  0.4 s, (a) construct the v t curve for 0  t  0.4 s, (b) determine the position of the part at t  0.3 s and t  0.2 s.

SOLUTION Divide the area of the a t curve into the four areas A1, A2 , A3 and A4. 2 (8)(0.2)  1.0667 ft/s 3 A2  (16)(0.2)  3.2 ft/s A1 

1 (16  8)(0.1)  1.2 ft/s 2 1 A4  (8)(0.1)  0.4 ft/s 2 A3 

Velocities: v0  0 v0.2  v0  A1  A2

v0.2  4.27 ft/s 

v0.3  v0.2  A3

v0.3  5.47 ft/s 

v0.4  v0.3  A4

v0.4  5.87 ft/s 

Sketch the v t curve and divide its area into A5 , A6 , and A7 as shown. 0.8 0.4  x dx  0.8  x   t vdt

At t  0.3 s, With A5 

With A5  A6 

0.4

x  0.8   t vdt

x0.3  0.8  A5  (5.47)(0.1)

2 (0.4)(0.1)  0.0267 ft, 3

At t  0.2 s,

and

or

x0.3  0.227 ft 

x0.2  0.8  ( A5  A6 )  A7 2 (1.6)(0.2)  0.2133 ft, 3

A7  (4.27)(0.2)  0.8533 ft

x0.2  0.8  0.2133  0.8533

x0.2  0.267 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.78 A car is traveling at a constant speed of 54 km/h when its driver sees a child run into the road. The driver applies her brakes until the child returns to the sidewalk and then accelerates to resume her original speed of 54 km/h; the acceleration record of the car is shown in the figure. Assuming x  0 when t  0, determine (a) the time t1 at which the velocity is again 54 km/h, (b) the position of the car at that time, (c) the average velocity of the car during the interval 1 s  t  t1.

SOLUTION Given:

At

t  0, x  0, v  54 km/h;

For t  t1 ,

v  54 km/h

First note (a)

54 km/h  15 m/s vb  va  (area under at curve from ta to tb )

We have Then

at

t  2 s:

v  15  (1)(6)  9 m/s

t  4.5 s:

1 v  9  (2.5)(6)  1.5 m/s 2

t  t1:

15  1.5 

1 (t1  4.5)(2) 2

or (b)

Using the above values of the velocities, the vt curve is drawn as shown.

t1  18.00 s 

PROBLEM 11.78 (Continued)

Now

x at t  18 s x18  0   (area under the v t curve from t  0 to t  18 s)  15  9   (1 s)(15 m/s)  (1 s)   m/s  2  1   + (2.5 s)(1.5 m/s)+ (2.5 s)(7.5 m/s)  3   2     (13.5 s)(1.5 m/s)  (13.5 s)(13.5 m/s)  3    [15  12  (3.75  6.25)  (20.25  121.50)] m

 178.75 m (c)

First note

or

x18  178.8 m 

x1  15 m x18  178.75 m

Now or

vave 

 x (178.75  15) m   9.6324 m/s t (18  1) s vave  34.7 km/h 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.79 An airport shuttle train travels between two terminals that are 1.6 mi apart. To maintain passenger comfort, the acceleration of the train is limited to 4 ft/s2, and the jerk, or rate of change of acceleration, is limited to 0.8 ft/s2 per second. If the shuttle has a maximum speed of 20 mi/h, determine (a) the shortest time for the shuttle to travel between the two terminals, (b) the corresponding average velocity of the shuttle.

SOLUTION xmax  1.6 mi; | amax|  4 ft/s 2

Given:

 da   0.8 ft/s2 /s; vmax  20 mi/h  dt   max First note

20 mi/h  29.333 ft/s 1.6 mi  8448 ft

(a)

To obtain tmin , the train must accelerate and decelerate at the maximum rate to maximize the time for which v  vmax . The time  t required for the train to have an acceleration of 4 ft/s2 is found from

a  da   max  dt  t  max 4 ft/s 2 0.8 ft/s 2 /s

or

t 

or

t  5 s

Now,

da    constant   since dt  

after 5 s, the speed of the train is

v5 

1 (t )(amax ) 2

or

v5 

1 (5 s)(4 ft/s 2 )  10 ft/s 2

Then, since v5  vmax , the train will continue to accelerate at 4 ft/s2 until v  vmax . The at curve must then have the shape shown. Note that the magnitude of the slope of each inclined portion of the curve is 0.8 ft/s2/s.

PROBLEM 11.79 (Continued)

Now

at t  (10  t1 ) s, v  vmax :

1  2  (5 s)(4 ft/s2 )   (t1 )(4 ft/s 2 )  29.333 ft/s 2  or

t1  2.3333 s

Then

at t  5 s: t  7.3333 s:

1 (5)(4)  10 ft/s 2 v  10  (2.3333)(4)  19.3332 ft/s

v 0

t  12.3333 s: v  19.3332 

1 (5)(4)  29.3332 ft/s 2

Using symmetry, the vt curve is then drawn as shown.

Noting that A1  A2  A3  A4 and that the area under the vt curve is equal to xmax, we have   10  19.3332   2 (2.3333 s)   ft/s  2      (10  t2 ) s  (29.3332 ft/s)  8448 ft

or

t2  275.67 s

Then

tmin  4(5 s)  2(2.3333 s)  275.67 s

 300.34 s tmin  5.01 min 

or (b)

We have or

vave 

x 1.6 mi 3600 s   t 300.34 s 1h

vave  19.18 mi/h 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.80 During a manufacturing process, a conveyor belt starts from rest and travels a total of 1.2 ft before temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to 4.8 ft/s2 per second, determine (a) the shortest time required for the belt to move 1.2 ft, (b) the maximum and average values of the velocity of the belt during that time.

SOLUTION Given:

(a)

At

t  0, x  0, v  0; xmax  1.2 ft;

when

 da  x  xmax , v  0;    4.8 ft/s 2  dt max

Observing that vmax must occur at t  12 tmin , the at curve must have the shape shown. Note that the magnitude of the slope of each portion of the curve is 4.8 ft/s2/s.

We have

at

t  t :

1 1 v  0  (t )(amax )  amax t 2 2

t  2t :

vmax 

1 1 amax t  (t )(amax )  amax t 2 2

Using symmetry, the vt is then drawn as shown.

Noting that A1  A2  A3  A4 and that the area under the vt curve is equal to xmax, we have (2t )(vmax )  xmax vmax  amax t  2amax t 2  xmax

PROBLEM 11.80 (Continued)

Now

or Then

amax  4.8 ft/s 2 /s so that t

2(4.8t ft/s3 )t 2  1.2 ft t  0.5 s tmin  4t tmin  2.00 s 

or (b)

We have

vmax  amax t  (4.8 ft/s 2 /s  t)t  4.8 ft/s 2 /s  (0.5 s) 2

vmax  1.2 ft/s 

or Also or

vave 

x 1.2 ft  ttotal 2.00 s vave  0.6 ft/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.81 Two seconds are required to bring the piston rod of an air cylinder to rest; the acceleration record of the piston rod during the 2 s is as shown. Determine by approximate means (a) the initial velocity of the piston rod, (b) the distance traveled by the piston rod as it is brought to rest.

SOLUTION

at curve; at

Given: 1.

t  2 s, v  0

The at curve is first approximated with a series of rectangles, each of width t  0.25 s. The area (t)(aave) of each rectangle is approximately equal to the change in velocity v for the specified interval of time. Thus, v  aave t where the values of aave and v are given in columns 1 and 2, respectively, of the following table.

2.

v(2)  v0 

Now and approximating the area



2 0



2 0

a dt  0

a dt under the at curve by aave t  v, the initial velocity is then

equal to v0  v Finally, using v2  v1  v12 where v12 is the change in velocity between times t1 and t2, the velocity at the end of each 0.25 interval can be computed; see column 3 of the table and the vt curve. 3.

The vt curve is then approximated with a series of rectangles, each of width 0.25 s. The area (t )(vave ) of each rectangle is approximately equal to the change in position  x for the specified interval of time. Thus  x  vave t where vave and  x are given in columns 4 and 5, respectively, of the table.

PROBLEM 11.81 (Continued)

4.

With x0  0 and noting that

x2  x1   x12 where  x12 is the change in position between times t1 and t2, the position at the end of each 0.25 s interval can be computed; see column 6 of the table and the xt curve.

(a)

We had found

(b)

At t  2 s

v0  1.914 m/s  x  0.840 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.82 The acceleration record shown was obtained during the speed trials of a sports car. Knowing that the car starts from rest, determine by approximate means (a) the velocity of the car at t  8 s, (b) the distance the car has traveled at t  20 s.

SOLUTION Given: at curve; at 1.

t  0, x  0, v  0

The at curve is first approximated with a series of rectangles, each of width t  2 s. The area (t )(aave ) of each rectangle is approximately equal to the change in velocity v for the specified interval of time. Thus, v  aave t where the values of aave and v are given in columns 1 and 2, respectively, of the following table.

2.

Noting that v0  0 and that v2  v1  v12 where v12 is the change in velocity between times t1 and t2, the velocity at the end of each 2 s interval can be computed; see column 3 of the table and the vt curve.

3.

The vt curve is next approximated with a series of rectangles, each of width t  2 s. The area (t )(vave ) of each rectangle is approximately equal to the change in position x for the specified interval of time. Thus,

x  vave t

where vave and x are given in columns 4 and 5, respectively, of the table. 4.

With x0  0 and noting that

x2  x1   x12 where  x12 is the change in position between times t1 and t2, the position at the end of each 2 s interval can be computed; see column 6 of the table and the xt curve.

PROBLEM 11.82 (Continued)

(a)

At t  8 s, v  32.58 m/s

(b)

At t  20 s

or

v  117.3 km/h 

x  660 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.83 A training airplane has a velocity of 126 ft/s when it lands on an aircraft carrier. As the arresting gear of the carrier brings the airplane to rest, the velocity and the acceleration of the airplane are recorded; the results are shown (solid curve) in the figure. Determine by approximate means (a) the time required for the airplane to come to rest, (b) the distance traveled in that time.

SOLUTION Given: a  v curve:

v0  126 ft/s The given curve is approximated by a series of uniformly accelerated motions (the horizontal dashed lines on the figure). For uniformly accelerated motion v22  v12  2a ( x2  x1 ) v2  v1  a (t2  t1 )

v22  v12 2a v v t  2 1 a

x 

or

For the five regions shown above, we have Region

v1 , ft/s

v2 , ft/s

a, ft/s 2

x, ft

t , s

1

126

120

12.5

59.0

0.480

2

120

100

33

66.7

0.606

3

100

80

45.5

39.6

0.440

4

80

40

54

44.4

0.741

5

40

0

58

13.8

0.690

223.5

2.957



(a)

From the table, when v  0

t  2.96 s 

(b)

From the table and assuming x0  0, when v  0

x  224 ft 

PROBLEM 11.84 Shown in the figure is a portion of the experimentally determined v  x curve for a shuttle cart. Determine by approximate means the acceleration of the cart (a) when x  10 in., (b) when v  80 in./s.

SOLUTION Given: v  x curve

First note that the slope of the above curve is

dv . dx

av (a)

Now

dv dx

When

x  10 in., v  55 in./s

Then

 40 in./s  a  55 in./s    13.5 in.  a  163.0 in./s 2 

or (b)

When v  80 in./s, we have

 40 in./s  a  80 in./s    28 in.  or

a  114.3 in./s 2 

Note: To use the method of measuring the subnormal outlined at the end of Section 11.8, it is necessary that the same scale be used for the x and v axes (e.g., 1 in.  50 in., 1 in.  50 in./s). In the above solution, v and  x were measured directly, so different scales could be used.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.85 An elevator starts from rest and rises 40 m to its maximum velocity in T s with the acceleration record shown in the figure. Determine (a) the required time T, (b) the maximum velocity, (c) the velocity and position of the elevator at t = T/2.

SOLUTION Find the position of the elevator from the a  t curve using the moment-area method. First, find the areas A1 and A2 on the a  t curve

A1 

1 T  0.6   0.1T m/s 2 3

A2 

1 2T  0.6   0.2T m/s 2 3

(a) Apply the moment-area formula: x  x0  v0t 

T 0 T  t  adt

   T 1  2T    2  T  x  v0t   A1   T      A2  T        3  3    3 3  3   7 2 8 2 40  0  T  T 90 90 15 2 T  90 1  T2 6 T 2   40  6   240 s 2

T  15.49 s  (b)

vmax  v0  A1  A2  0  0.1T  0.2T  0.3T vmax  4.65 m/s 

(c)

To find the position at t=T/2, first find the areas A1 and A2 on the a  t curve

A1  0.1T A4 

A3 

T 1  0.6   0.05T 2 6

1 T  0.45  0.0375T 2 6

Apply the moment-area formula: x  x0  v0t 

T 2  0  2  t  adt  

T

PROBLEM 11.85 (Continued)

x  v0

T  T 2T  2 T  1 T   A1     A3     A4    2 9  2 3 6 3 6

T  5T  T   0   0.1T      0.05T      0.0375T   0.035417T 2 18  18  9

  0.035417 15.49 

2

x  8.50 m  v  v0  A1  A3  A4  0.1875T

v  2.90 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.86 Two road rally checkpoints A and B are located on the same highway and are 8 mi apart. The speed limits for the first 5 mi and the last 3 mi are 60 mi/h and 35 mi/h, respectively. Drivers must stop at each checkpoint, and the specified time between points A and B is 10 min 20 s. Knowing that a driver accelerates and decelerates at the same constant rate, determine the magnitude of her acceleration if she travels at the speed limit as much as possible.

SOLUTION Given

10 20  60 3600  0.1722 h

10 min 20 s 

Sketch the v  t curve and note ta 

60 25 35 , tb  , tc  a a a

Find Areas A1 and A2 1 1  60   ta    25 tb 2 2 1 1  60 t1  1800  312.5 a a

(1)

1  35 tc 2 1  6.0278  35t1  612.5 a

(2)

A1  60t1 

A2  35  0.1722  t1  

Also given

A1  5 mi and A2  3 mi

Setting (1) equal to 5 mi:

60t1  2112.5

Setting (2) equal to 3 mi:

35t1  612.5

Solving equations (3) and (4) for t1 and

1 5 a

(3)

1  3.0278 a

(4)

1 : a

t1  85.45  103 h  5.13 min 1  60.23  106 h 2 / mi a

16.616  10   5280  3

3

a  16.616  10 mi/h

2

 3600 

2

a  6.77 ft/s 2 

PROBLEM 11.87 As shown in the figure, from t  0 to t  4 s, the acceleration of a given particle is represented by a parabola. Knowing that x  0 and v  8 m/s when t  0, (a) construct the v t and x–t curves for 0  t  4 s, (b) determine the position of the particle at t  3 s. (Hint: Use table inside the front cover.)

SOLUTION Given

At

t  0, x  0, v  8 m/s

(a) We have

v2  v1  (area under at curve from t1 to t2)

and

x2  x1  (area under vt curve from t1 to t2)

Then, using the formula for the area of a parabolic spandrel, we have 1 at t  2 s: v  8  (2)(12)  0 3 1 t  4 s: v  0  (2)(12)  8 m/s 3

The vt curve is then drawn as shown.

Note: The area under each portion of the curve is a spandrel of Order no. 3. (2)(8) 4m 3 1 (2)(8) t  4 s: x  4  0 3 1

Now at t  2 s: x  0 

The xt curve is then drawn as shown. (b) We have

or



At t  3 s: a  3(3  2)2  3 m/s 2 1 v  0  (1)(3)  1 m/s 3 (1)(1) x 4 3 1

x3  3.75 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.88 A particle moves in a straight line with the acceleration shown in the figure. Knowing that the particle starts from the origin with v0  2 m/s, (a) construct the v  t and x  t curves for 0  t  18 s, (b) determine the position and the velocity of the particle and the total distance traveled when t  18 s.

SOLUTION Find the indicated areas under the a t curve: A1   0.758  6 m/s A2   2 4  8 m/s A3   6 6  36 m/s

Find the values of velocity at times indicated: v0  2 m/s v8  v0  A1  8 m/s v12  v8  A2  0

v18  v12  A3  36 m/s (a) Sketch v  t curve using straight line portions over the constant acceleration periods. Find the indicated areas under the v  t curve: 1  2  88  40 m 2 1 A5   8 4  16 m 2 1 A6   36 6  108 m 2

A4 

Find the values of position at times indicated: x0  0 x8  x0  A4  40 m x12  x8  A5  56 m

x18  x12  A6  52 m Sketch the x  t curve using the positions and times calculated: (b) Position at t=18s Velocity at t-18s Total Distance Traveled:  56  108

x18  52 m  v18  36 m/s  d  164 m 

PROBLEM 11.89 A ball is thrown so that the motion is defined by the equations x  5t and y  2  6t  4.9t 2 , where x and y are expressed in meters and t is expressed in seconds. Determine (a) the velocity at t  1 s, (b) the horizontal distance the ball travels before hitting the ground.

SOLUTION Units are meters and seconds. Horizontal motion:

vx 

dx 5 dt

Vertical motion:

vy 

dy  6  9.8t dt

(a)

Velocity at t  1 s.

vx  5 v y  6  9.8  3.8

v  vx2  v 2y  52  3.82  6.28 m/s tan  

(b)

vy vx



3.8 5

Horizontal distance:

  37.2

v  6.28 m/s

37.2 

( y  0) y  2  6t  4.9t 2 t  1.4971 s x  (5)(1.4971)  7.4856 m

x  7.49 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.90 The motion of a vibrating particle is defined by the position vector r  10(1  e3t )i  (4e2t sin15t ) j, where r and t are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when (a) t  0, (b) t  0.5 s.

SOLUTION r  10(1  e3t )i  (4e2t sin15t ) j Then

v

and

a

(a)

dr  30e 3t i  [60e 2t cos15t  8e 2t sin15t ]j dt

dv  90e3t i  [120e2t cos15t  900e2t sin15t  120e2t cos15t  16e 2t sin15t ]j dt  90e3t i  [240e2t cos15t  884e2t sin15t ]j

When t  0:

v  30i  60 j mm/s

v  67.1 mm/s

63.4 

a  90i  240 j mm/s 2

a  256 mm/s 2

69.4 

v  8.29 mm/s

36.2 

a  336 mm/s2

86.6 

When t  0.5 s:

v  30e1.5 i  [60e1 cos 7.5  8e1 sin 7.5]  6.694i  4.8906 j mm/s

a  90e1.5i  [240e1 cos 7.5  884e1 sin 7.5 j]  20.08i  335.65 j mm/s2

PROBLEM 11.91 The motion of a vibrating particle is defined by the position vector r  (4sin  t )i  (cos 2 t ) j, where r is expressed in inches and t in seconds. (a) Determine the velocity and acceleration when t  1 s. (b) Show that the path of the particle is parabolic.

SOLUTION r  (4sin  t )i  (cos 2 t ) j v  (4 cos  t )i  (2 sin 2 t ) j a  (4 2 sin  t )i  (4 2 cos 2 t ) j

(a)

When t  1 s: v  (4 cos  )i  (2 sin 2 ) j

v  (4 in/s)i 

a  (4 2 sin  )i  (4 2 cos  ) j (b)

a  (4 2in/s2 ) j 

Path of particle: Since r  xi  y j ;

x  4sin  t ,

y   cos 2 t

Recall that cos 2  1  2sin 2  and write

y   cos 2 t  (1  2sin 2  t ) But since x  4sin  t or sin  t 

(1)

1 x, Eq.(1) yields 4

2  1   y   1  2  x    4   

y

1 2 x  1 (Parabola)  8

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.92 The motion of a particle is defined by the equations x  10t  5sin t and y  10  5cos t , where x and y are expresed in feet and t is expressed in seconds. Sketch the path of the particle for the time interval 0  t  2 , and determine (a) the magnitudes of the smallest and largest velocities reached by the particle, (b) the corresponding times, positions, and directions of the velocities.

SOLUTION Sketch the path of the particle, i.e., plot of y versus x. Using x  10t  5sin t , and y  10  5cos t obtain the values in the table below. Plot as shown. t(s)

x(ft)

y(ft)

0

0.00

5

 2

10.71

10



31.41

15

 2

52.12

10

2

62.83

5

3

PROBLEM 11.92 (Continued)

(a)

Differentiate with respect to t to obtain velocity components. vx 

dx  10  5cos t and v y  5sin t dt

v 2  vx2  v 2y  (10  5cos t )2  25sin 2 t  125  100cos t d (v ) 2  100sin t  0 t  0.   .  2   N  dt When t  2 N  .

cos t  1.

and

v2 is minimum.

When t  (2 N  1) .

cos t  1.

and

v2 is maximum.

(v 2 )min  125  100  25(ft/s)2 vmin  5 ft/s  (v 2 )max  125  100  225(ft/s)2 (b)

vmax  15 ft/s 

When v  vmin. When N  0,1, 2,

x  10(2 N )  5sin(2 N )

x  20 N ft 

y  10  5cos(2 N )

y  5 ft 

vx  10  5cos(2 N )

vx  5 ft/s 

v y  5sin(2 N ) tan  

vy vx

 0,

vy  0 

 0 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.92 (Continued)

t  (2 N  1) s 

When v  vmax . x  10[2 ( N  1)]  5sin[2 ( N  1)]

x  20 ( N  1) ft 

y  10  5cos[2 ( N  1)]

y  15 ft 

vx  10  5cos[2 ( N  1)]

vx  15 ft/s 

v y  5sin[2 ( N  1)] tan  

vy vx

 0,

vy  0 

 0 

PROBLEM 11.93 The damped motion of a vibrating particle is defined by the position vector r  x1[1  1/(t  1)]i  ( y1e t/ 2 cos 2 t ) j, where t is expressed in seconds. For x1  30 mm and y1  20 mm, determine the position, the velocity, and the acceleration of the particle when (a) t  0, (b) t  1.5 s.

SOLUTION We have

1     t/ 2 cos 2 t ) j r  30  1   i  20(e  t 1

Then

v

 30

1    i  20   e t/ 2 cos 2 t  2 e t/ 2 sin 2 t  j 2 (t  1)  2 

 30

 1 1  i  20 e t/ 2  cos 2 t  2 sin 2 t   j 2 2 (t  1)   

a

and

dr dt

dv dt

 30 

(a)

At t  0:

   2 1  i  20   e t/2  cos 2 t  2 sin 2 t   e t/ 2 ( sin 2 t  4 cos 2 t )  j 3 (t  1) 2   2 

60 i  10 2 e t/ 2 (4 sin 2 t  7.5 cos 2 t ) j (t  1)3

 1 r  30 1   i  20(1) j  1 r  20 mm 

or

 1 1  v  30   i  20 (1)   0   j 1   2 v  43.4 mm/s

or

46.3° 

60 a   i  10 2 (1)(0  7.5) j (1) or

a  743 mm/s 2

85.4° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.93 (Continued)

(b)

At t  1.5 s:

1   0.75 r  30 1  (cos 3 ) j  i  20e 2.5    (18 mm)i  (1.8956 mm) j

or

r  18.10 mm

6.01° 

v  5.65 mm/s

31.8° 

a  70.3 mm/s 2

86.9° 

30 1  i  20 e0.75  cos 3  0  j 2 2 (2.5)    (4.80 mm/s)i  (2.9778 mm/s) j

v

or a

60 i  10 2 e 0.75 (0  7.5 cos 3 ) j (2.5)3

 ( 3.84 mm/s 2 )i  (70.1582 mm/s 2 ) j

or

Problem 11.94 A girl operates a radio-controlled model car in a vacant parking lot. The girl’s position is at the origin of the xy coordinate axes, and the surface of the parking lot lies in the x-y plane. The motion of the car is defined by the  position vector r  (2  2t 2 )iˆ  (6  t 3 ) ˆj where r and t are expressed in meters and seconds, respectively. Determine (a) the distance between the car and the girl when t = 2 s, (b) the distance the car traveled in the interval from t = 0 to t = 2 s, (c) the speed and direction of the car’s velocity at t = 2 s, (d) the magnitude of the car’s acceleration at t = 2 s.

SOLUTION Given:

r  (2  2t 2 )i  (6  t 3 ) j

(a) At t=2s

r  2   10i  14 j m r  2   10 2  14 2 m

r  2  =17.20 m 

(b) The car is traveling on a curved path. The distance that the car travels during any infinitesimal interval is given by:

ds  dx 2  dy 2

where:

dx  4tdt and dy  3t 2 dt

Substituting

ds  16t 2  9t 4 dt

Integrating:



(c) Velocity can be found by

s

0

2

ds   t 16  9t 2 dt 0

s

3/2 2 1 16  9t 2  |  0 27

v

dr dt

s  11.52 m 

v  4ti  3t 2 j m/s At t=2 s

v  8i  12 j m/s

(d) Acceleration can be found by a 

v  14.42 m/s

56.31 

dv dt

a  4i  6tj m/s 2 At t = 2 s

a  4i  12 j m/s 2

a  12.65 m/s 2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r  (Rt cos nt)i  ctj  (Rt sin nt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.)

SOLUTION We have

r  ( Rt cos n t )i  ctj  ( Rt sin n t )k

Then

v

and

a

dr  R(cos n t  n t sin n t )i  cj  R(sin n t  n t cos n t )k dt

dv dt  R (n sin n t  n sin n t  n2 t cos n t )i  R (n cos n t  n cos n t  n2t sin n t )k  R (2n sin n t  n2 t cos n t )i  R(2n cos n t  n2 t sin n t )k

Now

v 2  vx2  v 2y  vz2  [ R(cos n t  n t sin n t )]2  (c)2  [ R (sin n t  n t cos n t )]2





 R 2  cos 2 n t  2n t sin n t cos n t  n2 t 2 sin 2 n t   sin 2 n t  2n t sin n t cos n t  n2 t 2 cos 2 n t   c 2 









 R 2 1  n2t 2  c 2



Also,



v  R 2 1  n2 t 2  c 2 

or

a 2  ax2  a y2  az2





2

  R 2n sin n t  n2 t cos n t   (0) 2  

 



2

  R 2n cos n t  n2 t sin n t    2 2 2 3  R 4n sin n t  4n t sin n t cos n t  n4t 2 cos 2 n t 







 4n2 cos 2 n t  4n3t sin n t cos n t  n4t 2 sin 2 n t  



 R 2 4n2  n4 t 2

or



a  Rn 4  n2t 2 

PROBLEM 11.96 The three-dimensional motion of a particle is defined by the position vector r  ( At cos t )i  ( A t 2  1) j  ( Bt sin t )k , where r and t are expressed in feet and seconds, respectively. Show that the curve described by the particle lies on the hyperboloid (y/A)2  (x/A)2  (z/B)2  1. For A  3 and B  1, determine (a) the magnitudes of the velocity and acceleration when t  0, (b) the smallest nonzero value of t for which the position vector and the velocity are perpendicular to each other.

SOLUTION We have

r  ( At cos t )i  ( A t 2  1) j  ( Bt sin t )k

or

x  At cos t cos t 

Then

x At

y  A t 2  1 z  Bt sin t sin t 

2

 y t2    1  A

z Bt 2

2

2

2

2

x z t2        A  B

or 2

Then

 y x z  A  1   A    B       

or

 y x  z  A    A   B  1      

2

(a)

2

 x   z  cos 2 t  sin 2 t  1        1  At   Bt 

Now

2

2



Q.E.D.

With A  3 and B  1, we have

v

dr t  3(cos t  t sin t )i  3 j  (sin t  t cos t )k 2 dt t 1

  t

and

t 2  1  t t 2 1 dv a j  3( sin t  sin t  t cos t )i  3 dt (t 2  1)  (cos t  cos t  t sin t )k 1 j  (2 cos t  t sin t )k  3(2 sin t  t cos t )i  3 2 (t  1)3/2

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.96 (Continued)

At t  0:

v  3(1  0)i  (0) j  (0)k

v  vx2  v 2y  vz2 v  3 ft/s 

or and Then

a  3(0)i  3(1) j  (2  0)k

a 2  (0)2  (3)2  (2)2  13 a  3.61 ft/s 2 

or (b)

If r and v are perpendicular, r  v  0 t   [(3t cos t )i  (3 t 2  1) j  (t sin t )k ]  [3(cos t  t sin t )i   3  j  (sin t  t cos t )k ]  0 2  t 1 

or Expanding

t   (3t cos t )[3(cos t  t sin t )]  (3 t 2  1)  3   (t sin t )(sin t  t cos t )  0 2  t 1 

(9t cos2 t  9t 2 sin t cos t )  (9t )  (t sin 2 t  t 2 sin t cos t )  0

or (with t  0)

10  8 cos2 t  8t sin t cos t  0

or Using “trial and error” or numerical methods, the smallest root is Note: The next root is t  4.38 s. 

7  2 cos 2t  2t sin 2t  0 t  3.82 s 

PROBLEM 11.97 An airplane used to drop water on brushfires is flying horizontally in a straight line at 180 mi/h at an altitude of 300 ft. Determine the distance d at which the pilot should release the water so that it will hit the fire at B.

SOLUTION First note

v0  180 km/h  264 ft/s

Place origin of coordinates at Point A. Vertical motion. (Uniformly accelerated motion) y  0  (0)t 

1 2 gt 2

1 300 ft   (32.2 ft/s 2 )t 2 2

At B:

t B  4.31666 s

or Horizontal motion. (Uniform)

x  0  (vx )0 t At B: or

d  (264 ft/s)(4.31666 s) d  1140 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.98 A ski jumper starts with a horizontal take-off velocity of 25 m/s and lands on a straight landing hill inclined at 30o. Determine (a) the time between take-off and landing, (b) the length d of the jump, (c) the maximum vertical distance between the jumper and the landing hill.

SOLUTION y   x tan 30

(a) At the landing point,

x  x0   v x 0 t  v0t

Horizontal motion:

 0 t  12 gt 2   12 gt 2

Vertical motion:

y  y0  v y

from which

t2  

2y 2 x tan 30 2v0t tan 30   g g g

t 

Rejecting the t  0 solution gives

d 

(b) Landing distance:

2v0 tan 30  2  25  tan 30  9.81 g

 25 2.94  x v0t   cos 30 cos 30 cos 30

t  2.94 s 

d  84.9 m 

h  x tan 30  y

(c) Vertical distance:

h  v0t tan 30 

or

1 2 gt 2

Differentiating and setting equal to zero, dh  v0 tan 30  gt  0 dt Then,

hmax 

or

t 

vo tan 30 g

 v0  v0 tan 30 tan 30  1 g  v0 tan 30   2 

g

v 2 tan 2 30  25  tan 30   0  2g  2 9.81 2

g

2

 

2

hmax  10.62 m 

PROBLEM 11.99 A baseball pitching machine “throws” baseballs with a horizontal velocity v0. Knowing that height h varies between 788 mm and 1068 mm, determine (a) the range of values of v0, (b) the values of  corresponding to h  788 mm and h  1068 mm.

SOLUTION

(a)

y0  1.5 m, (v y )0  0

Vertical motion:

or

t 

2( y0  y) g

y h

or

tB 

2( y0  h) g

When h  788 mm  0.788 m,

tB 

(2)(1.5  0.788)  0.3810 s 9.81

When h  1068 mm  1.068 m,

tB 

(2)(1.5  1.068)  0.2968 s 9.81

y  y0  (v y )0 t  At Point B,

Horizontal motion:

1 2 gt 2

x0  0, (vx )0  v0 , x  v0t

or

v0 

x x  B t tB

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.99 (Continued)

v0 

12.2  32.02 m/s 0.3810

and

v0 

12.2  41.11 m/s 0.2968

32.02 m/s  v0  41.11 m/s

or

Vertical motion:

v y  (v y )0  gt   gt

Horizontal motion:

vx  v0

With xB  12.2 m,

(b)

we get

tan   

115.3 km/h  v0  148.0 km/h 

(v y ) B dy gt   B dx (vx ) B v0

For h  0.788 m,

tan  

(9.81)(0.3810)  0.11673, 32.02

  6.66 

For h  1.068 m,

tan  

(9.81)(0.2968)  0.07082, 41.11

  4.05 

PROBLEM 11.100 While delivering newspapers, a girl throws a newspaper with a horizontal velocity v0. Determine the range of values of v0 if the newspaper is to land between Points B and C.

SOLUTION Vertical motion. (Uniformly accelerated motion) y  0  (0)t 

1 2 gt 2

Horizontal motion. (Uniform)

x  0  (vx )0 t  v0t At B:

y:

tB  0.455016 s

or Then

x:

y:

or

1 2 ft   (32.2 ft/s 2 )t 2 2

tC  0.352454 s

or Then

7 ft  (v0 ) B (0.455016 s) (v0 ) B  15.38 ft/s

or At C:

1 1 3 ft   (32.2 ft/s 2 )t 2 3 2

x:

1 12 ft  (v0 )C (0.352454 s) 3

(v0 )C  35.0 ft/s 15.38 ft/s  v0  35.0 ft/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.101 Water flows from a drain spout with an initial velocity of 2.5 ft/s at an angle of 15° with the horizontal. Determine the range of values of the distance d for which the water will enter the trough BC.

SOLUTION First note

(vx )0  (2.5 ft/s) cos 15  2.4148 ft/s (v y )0  (2.5 ft/s) sin 15  0.64705 ft/s

Vertical motion. (Uniformly accelerated motion) y  0  (v y ) 0 t 

1 2 gt 2

At the top of the trough

1 8.8 ft  (0.64705 ft/s) t  (32.2 ft/s2 ) t 2 2 or

tBC  0.719491 s

(the other root is negative)

Horizontal motion. (Uniform)

x  0  (v x ) 0 t In time t BC

xBC  (2.4148 ft/s)(0.719491 s)  1.737 ft

Thus, the trough must be placed so that

xB  1.737 ft or xC  1.737 ft Since the trough is 2 ft wide, it then follows that

0  d  1.737 ft 

PROBLEM 11.102 In slow pitch softball the underhand pitch must reach a maximum height of between 1.8 m and 3.7 m above the ground. A pitch is made with an initial velocity v 0 of magnitude 13 m/s at an angle of 33 with the horizontal. Determine (a) if the pitch meets the maximum height requirement, (b) the height of the ball as it reaches the batter.

SOLUTION v0  13 m/s,   33, x0  0, y0  0.6 m v y  v0 sin   gt

Vertical motion:

y  y0   v0 sin   t 

At maximum height,

vy  0

(a)

t 

or

t 

1 2 gt 2 v0 sin  g

13sin 33  0.7217 s 9.81

ymax  0.6  13sin 33  0.7217  

1  9.81 0.7217 2 2

ymax  3.16 m 

1.8 m  3.16 m  3.7 m

Horizontal motion:

x  x0   v0 cos   t

At x  15.2 m,

t 

(b) Corresponding value of y :

yes  or

t 

x  x0 v0 cos 

15.2  0  1.3941 s 13cos 33

y  0.6  13sin 33 1.3941 

1  9.811.39412 2

y  0.937 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.103 A volleyball player serves the ball with an initial velocity v0 of magnitude 13.40 m/s at an angle of 20° with the horizontal. Determine (a) if the ball will clear the top of the net, (b) how far from the net the ball will land.

SOLUTION First note

(vx )0  (13.40 m/s) cos 20  12.5919 m/s (v y )0  (13.40 m/s) sin 20  4.5831 m/s

(a)

Horizontal motion. (Uniform)

x  0  (v x ) 0 t At C

9 m  (12.5919 m/s) t or tC  0.71475 s

Vertical motion. (Uniformly accelerated motion) y  y0  (v y )0 t  At C:

1 2 gt 2

yC  2.1 m  (4.5831 m/s)(0.71475 s) 1  (9.81 m/s2 )(0.71475 s)2 2  2.87 m

yC  2.43 m (height of net)  ball clears net  (b)

At B, y  0:

1 0  2.1 m  (4.5831 m/s)t  (9.81 m/s 2 )t 2 2

Solving

t B  1.271175 s (the other root is negative)

Then

d  (vx )0 tB  (12.5919 m/s)(1.271175 s)  16.01 m

The ball lands

b  (16.01  9.00) m  7.01 m from the net 

PROBLEM 11.104 A golfer hits a golf ball with an initial velocity of 160 ft/s at an angle of 25° with the horizontal. Knowing that the fairway slopes downward at an average angle of 5°, determine the distance d between the golfer and Point B where the ball first lands.

SOLUTION (vx )0  (160 ft/s) cos 25

First note

(v y )0  (160 ft/s) sin 25

xB  d cos 5

and at B Now

yB  d sin 5

Horizontal motion. (Uniform)

x  0  (v x ) 0 t d cos 5  (160 cos 25)t or t B 

At B

cos 5 d 160 cos 25

Vertical motion. (Uniformly accelerated motion)

y  0  (v y ) 0 t 

1 2 gt 2

( g  32.2 ft/s 2 ) 1 2 gt B 2

At B:

 d sin 5  (160 sin 25)t B 

Substituting for tB

 cos 5  1  cos 5  2 d sin 5  (160 sin 25)  d  g   d 2  160 cos 25   160 cos 25 

2

or

d

2 (160 cos 25) 2 (tan 5  tan 25) 32.2 cos 5

 726.06 ft or

d  242 yd 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.105 A homeowner uses a snowblower to clear his driveway. Knowing that the snow is discharged at an average angle of 40° with the horizontal, determine the initial velocity v0 of the snow.

SOLUTION First note

(vx )0  v0 cos 40 (v y )0  v0 sin 40

Horizontal motion. (Uniform)

x  0  (v x ) 0 t At B:

14  (v0 cos 40) t or tB 

14 v0 cos 40

Vertical motion. (Uniformly accelerated motion) y  0  (v y ) 0 t  At B:

1 2 gt 2

1.5  (v0 sin 40) t B 

( g  32.2 ft/s 2 )

1 2 gt B 2

Substituting for tB   1   14 14 1.5  (v0 sin 40)   g   v0 cos 40  2  v0 cos 40 

or or

v02 

1 2

2

(32.2)(196)/ cos 2 40 1.5  14 tan 40 v0  22.9 ft/s 

PROBLEM 11.106 At halftime of a football game souvenir balls are thrown to the spectators with a velocity v0. Determine the range of values of v0 if the balls are to land between Points B and C.

SOLUTION The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below Point A. The coordinates of Point A are x0  0, y0  2m. The components of initial velocity are (vx )0  v0 cos 40 m/s and (v y )0  v0 sin 40. Horizontal motion:

x  x0  (vx )0 t  (v0 cos 40)t

Vertical motion:

y  y0  (v y )0 t 

(1)

1 2 gt 2

1  2  (v0 sin 40)   (9.81)t 2 2

From (1),

v0t 

(2)

x cos 40

(3)

y  2  x tan 40  4.905t 2

Then

t2 

Point B:

2  x tan 40  y 4.905

(4)

x  8  10cos 35  16.1915 m y  1.5  10sin 35  7.2358 m 16.1915  21.1365 m cos 40 2  16.1915 tan 40  7.2358 t2  4.905 21.1365 v0  1.3048

v0 t 

t  1.3048 s v0  16.199 m/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.106 (Continued)

Point C:

x  8  (10  7) cos 35  21.9256 m y  1.5  (10  7)sin 35  11.2508 m

Range of values of v0.

v0 t 

21.9256  28.622 m cos 40

t2 

2  21.9256 tan 40  11.2508 4.905

v0 

28.622 1.3656

t  1.3656 s

v0  20.96 m/s 16.20 m/s  v0  21.0 m/s 

PROBLEM 11.107 A basketball player shoots when she is 16 ft from the backboard. Knowing that the ball has an initial velocity v0 at an angle of 30° with the horizontal, determine the value of v0 when d is equal to (a) 9 in., (b) 17 in.

SOLUTION First note

(vx )0  v0 cos 30

Horizontal motion. (Uniform)

(v y )0  v0 sin 30

x  0  (v x ) 0 t

(16  d )  (v0 cos 30) t or t B 

At B:

16  d v0 cos 30

y  0  (v y ) 0 t 

Vertical motion. (Uniformly accelerated motion)

1 2 gt 2

1 2 gt B 2

At B:

3.2  (v0 sin 30) t B 

Substituting for tB

 16  d  1  16  d  3.2  (v0 sin 30)   g   v0 cos 30  2  v0 cos 30 

or

v02 

(a)

d  9 in.:

v02

d  17 in.:

v02

(b)





( g  32.2 ft/s 2 )

2

2 g (16  d ) 2 3 

1 3

(16  d )  3.2  

2(32.2) 16  129  3 

1 3

16  129   3.2

2(32.2) 16  17 12  3 

1 3

2

v0  29.8 ft/s 

2

16  1712   3.2

v0  29.6 ft/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.108 A tennis player serves the ball at a height h  2.5 m with an initial velocity of v0 at an angle of 5° with the horizontal. Determine the range for which of v0 for which the ball will land in the service area which extends to 6.4 m beyond the net.

SOLUTION The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below the point where the racket impacts the ball. The coordinates of this impact point are x0  0, y0  h  2.5 m. The components of initial velocity are (vx )0  v0 cos 5 and (v y )0  v0 sin 5. Horizontal motion:

x  x0  (vx )0 t  (v0 cos 5)t

Vertical motion:

y  y0  (v y )0 t 

(1)

1 2 gt 2

1  2.5  (v0 sin 5)t   (9.81)t 2 2

From (1), Then

v0t 

x cos 5

(2) (3)

y  2.5  x tan 5  4.905t 2 t2 

2.5  x tan 5  y 4.905

(4)

At the minimum speed the ball just clears the net. x  12.2 m, y  0.914 m 12.2 v0t   12.2466 m cos5 2.5  12.2 tan 5  0.914 t2  4.905 12.2466 v0  0.32517

t  0.32517 s v0  37.66 m/s

PROBLEM 11.108 (Continued)

At the maximum speed the ball lands 6.4 m beyond the net. x  12.2  6.4  18.6 m

y0

18.6  18.6710 m cos 5 2.5  18.6 tan 5  0 t2  t  0.42181 s 4.905 18.6710 v0  v0  44.26 m/s 0.42181

v0t 

Range for v0.

37.7 m/s  v0  44.3 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.109 The nozzle at A discharges cooling water with an initial velocity v0 at an angle of 6° with the horizontal onto a grinding wheel 350 mm in diameter. Determine the range of values of the initial velocity for which the water will land on the grinding wheel between Points B and C.

SOLUTION First note

(vx )0  v0 cos 6 (v y )0  v0 sin 6 Horizontal motion. (Uniform)

x  x0  (vx )0 t Vertical motion. (Uniformly accelerated motion) y  y0  (v y )0 t 

1 2 gt 2

( g  9.81 m/s 2 )

x  (0.175 m) sin 10 y  (0.175 m) cos 10

At Point B:

x : 0.175 sin 10  0.020  (v0 cos 6)t tB 

or

0.050388 v0 cos 6

y : 0.175 cos 10  0.205  (v0 sin 6)t B 

1 2 gt B 2

Substituting for tB  0.050388  1  0.050388  0.032659  (v0 sin 6)    (9.81)    v0 cos 6  2  v0 cos 6 

2

PROBLEM 11.109 (Continued)

v02

or



1 2

(9.81)(0.050388) 2

cos 2 6(0.032659  0.050388 tan 6)

(v0 ) B  0.678 m/s

or

x  (0.175 m) cos 30

At Point C:

y  (0.175 m) sin 30 x : 0.175 cos 30  0.020  (v0 cos 6)t tC 

or

0.171554 v0 cos 6

y : 0.175 sin 30  0.205  ( v0 sin 6)tC 

1 2 gtC 2

Substituting for tC  0.171554  1  0.171554  0.117500  ( v0 sin 6)    (9.81)    v0 cos 6  2  v0 cos 6 

or or



v02 

1 2

2

(9.81)(0.171554) 2

cos 2 6(0.117500  0.171554 tan 6)

(v0 )C  1.211 m/s 0.678 m/s  v0  1.211 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.110 While holding one of its ends, a worker lobs a coil of rope over the lowest limb of a tree. If he throws the rope with an initial velocity v0 at an angle of 65° with the horizontal, determine the range of values of v0 for which the rope will go over only the lowest limb.

SOLUTION First note (vx )0  v0 cos 65 (v y )0  v0 sin 65

Horizontal motion. (Uniform)

x  0  (v x ) 0 t At either B or C, x  5 m

s  (v0 cos 65)tB,C or

t B ,C 

5 (v0 cos 65)

Vertical motion. (Uniformly accelerated motion) y  0  (v y ) 0 t 

1 2 gt 2

( g  9.81 m/s 2 )

At the tree limbs, t  t B ,C   1   5 5 y B ,C  (v0 sin 65)   g  cos 65 2 cos 65 v v    0   0 

2

PROBLEM 11.110 (Continued)

v02 

or



1 2

(9.81)(25)

2

cos 65(5 tan 65  yB , C ) 686.566 5 tan 65  yB , C

At Point B:

v02 

686.566 5 tan 65  5

or

(v0 ) B  10.95 m/s

At Point C:

v02 

686.566 5 tan 65  5.9

or

(v0 )C  11.93 m/s



10.95 m/s  v0  11.93 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.111 The pitcher in a softball game throws a ball with an initial velocity v0 of 72 km/h at an angle  with the horizontal. If the height of the ball at Point B is 0.68 m, determine (a) the angle , (b) the angle  that the velocity of the ball at Point B forms with the horizontal.

SOLUTION First note

v0  72 km/h  20 m/s (vx )0  v0 cos   (20 m/s) cos 

and

(v y )0  v0 sin   (20 m/s) sin 

(a)

Horizontal motion. (Uniform)

x  0  (vx )0 t  (20 cos  ) t At Point B:

14  (20 cos  )t or tB 

7 10 cos 

Vertical motion. (Uniformly accelerated motion) y  0  (v y ) 0 t  At Point B:

1 2 1 gt  (20 sin  )t  gt 2 2 2

0.08  (20 sin  )t B 

( g  9.81 m/s 2 )

1 2 gt B 2

Substituting for tB

  1   7 7 0.08  (20 sin  )   g   10 cos   2  10 cos   or Now

8  1400 tan  

1 49 g 2 cos 2 

1  sec2   1  tan 2  cos 2 

2

PROBLEM 11.111 (Continued)

Then or Solving

8  1400 tan   24.5 g (1  tan 2  ) 240.345 tan 2   1400 tan   248.345  0

  10.3786 and   79.949

Rejecting the second root because it is not physically reasonable, we have

  10.38  (b)

We have

vx  (vx )0  20 cos 

and

v y  (v y )0  gt  20 sin   gt

At Point B:

(v y ) B  20 sin   gt B  20 sin  

7g 10 cos 

Noting that at Point B, v y  0, we have tan     or

|(v y ) B | vx 7g 10 cos 

 20 sin 

20 cos  7 9.81 200 cos 10.3786

 sin 10.3786

cos 10.3786

  9.74 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.112 A model rocket is launched from Point A with an initial velocity v0 of 75 m/s. If the rocket’s descent parachute does not deploy and the rocket lands a distance d  100 m from A, determine (a) the angle  that v0 forms with the vertical, (b) the maximum height above Point A reached by the rocket, and (c) the duration of the flight.

SOLUTION Set the origin at Point A.

x0  0,

y0  0

Horizontal motion:

x  v0 t sin 

Vertical motion:

y  v0 t cos  

cos  

sin 2   cos 2  

sin  

x v0 t

1 2 gt 2

1  1  y  gt 2   v0t  2 

(2)

2 1  2  1 2    x y gt     1 2 (v0t )2    

x 2  y 2  gyt 2 



1 24 g t  v02 t 2 4



1 24 g t  v02  gy t 2  ( x 2  y 2 )  0 4

At Point B,

(1)

x 2  y 2  100 m,

x  100 cos 30 m

y  100 sin 30  50 m

1 (9.81)2 t 4  [752  (9.81)(50)] t 2  1002  0 4 24.0590 t 4  6115.5 t 2  10000  0

t 2  252.54 s 2

and 1.6458 s 2

t  15.8916 s

and 1.2829 s

(3)

PROBLEM 11.112 (Continued)

Restrictions on  :

0    120

tan  

x y  gt 1 2

2



100 cos 30  0.0729 50  (4.905)(15.8916)2

  4.1669 100 cos 30  2.0655 50  (4.905)(1.2829) 2   115.8331

and

Use   4.1669 corresponding to the steeper possible trajectory. (a)

Angle  .

(b)

Maximum height.

  4.17  v y  0 at

y  ymax

v y  v0 cos   gt  0 t

v0 cos  g

ymax  v0 t cos   

(c)



Duration of the flight.

v 2 cos 2  1 gt  0 2 2g

(75) 2 cos 2 4.1669 (2)(9.81)

ymax  285 m 

(time to reach B) t  15.89 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.113 The initial velocity v0 of a hockey puck is 105 mi/h. Determine (a) the largest value (less than 45°) of the angle  for which the puck will enter the net, (b) the corresponding time required for the puck to reach the net.

SOLUTION First note

v0  105 mi/h  154 ft/s (vx )0  v0 cos   (154 ft/s) cos 

and

(v y )0  v0 sin   (154 ft/s) sin 

(a)

Horizontal motion. (Uniform)

x  0  (vx )0 t  (154 cos  )t At the front of the net, Then or

x  16 ft

16  (154 cos  )t tenter 

8 77 cos 

Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1  (154 sin  ) t  gt 2 2

y  0  (v y ) 0 t 

( g  32.2 ft/s 2 )

At the front of the net, yfront  (154 sin  ) tenter 

1 2 gtenter 2

  1   8 8  (154 sin  )   g   77 cos   2  77 cos   32 g  16 tan   5929 cos 2 

2

PROBLEM 11.113 (Continued)

Now Then or

1  sec2   1  tan 2  cos 2  yfront  16 tan  

tan 2  

32 g (1  tan 2  ) 5929

 5929  5929 tan   1  yfront   0 2g 32 g   5929 2g



    5929 2g 



2





1/ 2

  4 1  5929 y 32 g front   2

Then

tan  

or

 5929 2  5929 5929      1  tan   yfront    4  32.2  4  32.2   32  32.2  

or

tan   46.0326  [(46.0326)2  (1  5.7541 yfront )]1/ 2

1/ 2

Now 0  yfront  4 ft so that the positive root will yield values of   45 for all values of yfront. When the negative root is selected,  increases as yfront is increased. Therefore, for  max , set yfront  yC  4 ft

(b)

Then

tan   46.0326  [(46.0326)2  (1  5.7541  4)]1/ 2

or

 max  14.6604

We had found

or

 max  14.66 

8 77 cos  8  77 cos 14.6604

tenter 

tenter  0.1074 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.114 A worker uses high-pressure water to clean the inside of a long drainpipe. If the water is discharged with an initial velocity v0 of 11.5 m/s, determine (a) the distance d to the farthest Point B on the top of the pipe that the worker can wash from his position at A, (b) the corresponding angle .

SOLUTION First note

(vx )0  v0 cos   (11.5 m/s) cos  (v y )0  v0 sin   (11.5 m/s) sin 

By observation, dmax occurs when

ymax  1.1 m.

Vertical motion. (Uniformly accelerated motion) v y  (v y )0  gt  (11.5 sin  )  gt

y  ymax

When Then

at

1 2 gt 2 1  (11.5 sin  ) t  gt 2 2

y  0  (v y )0 t 

B , (v y ) B  0

(v y ) B  0  (11.5 sin  )  gt 11.5 sin  g

( g  9.81 m/s 2 )

or

tB 

and

yB  (11.5 sin  ) t B 

1 2 gt B 2

Substituting for tB and noting yB  1.1 m  11.5 sin   1  11.5 sin   1.1  (11.5 sin  )   g  g g   2   1 (11.5)2 sin 2   2g

or

sin 2  

2.2  9.81 11.52

  23.8265

2

PROBLEM 11.114 (Continued)

(a)

Horizontal motion. (Uniform) x  0  (vx )0 t  (11.5 cos  ) t At Point B: where Then or

(b)

From above

x  d max tB 

and t  t B

11.5 sin 23.8265  0.47356 s 9.81

d max  (11.5)(cos 23.8265)(0.47356) d max  4.98 m 

  23.8 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.115 An oscillating garden sprinkler which discharges water with an initial velocity v0 of 8 m/s is used to water a vegetable garden. Determine the distance d to the farthest Point B that will be watered and the corresponding angle  when (a) the vegetables are just beginning to grow, (b) the height h of the corn is 1.8 m.

SOLUTION First note

(vx )0  v0 cos   (8 m/s) cos  (v y )0  v0 sin   (8 m/s) sin 

Horizontal motion. (Uniform)

x  0  (vx )0 t  (8 cos  ) t At Point B: or

x  d : d  (8 cos  ) t d tB  8 cos 

Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1  (8 sin  ) t  gt 2 ( g  9.81 m/s 2 ) 2 1 0  (8 sin  ) t B  gt B2 2 y  0  (v y ) 0 t 

At Point B: Simplifying and substituting for tB

0  8 sin   d

or (a)

1  d  g  2  8 cos  

64 sin 2 g

(1)

When h  0, the water can follow any physically possible trajectory. It then follows from Eq. (1) that d is maximum when 2  90

  45 

or Then or

d

64 sin (2  45) 9.81

d max  6.52 m 

PROBLEM 11.115 (Continued)

(b)

Based on Eq. (1) and the results of Part a, it can be concluded that d increases in value as  increases in value from 0 to 45° and then d decreases as  is further increased. Thus, dmax occurs for the value of  closest to 45° and for which the water just passes over the first row of corn plants. At this row, xcom  1.5 m tcorn 

so that

1.5 8 cos 

Also, with ycorn  h, we have h  (8 sin  ) tcorn 

1 2 gtcorn 2

Substituting for tcorn and noting h  1.8 m,

 1.5  1  1.5  1.8  (8 sin  )   g   8 cos   2  8 cos   or Now Then or

1.8  1.5 tan  

2

2.25 g 128 cos 2 

1  sec2   1  tan 2  2 cos  1.8  1.5 tan  

2.25(9.81) (1  tan 2  ) 128

0.172441 tan 2   1.5 tan   1.972441  0

  58.229 and   81.965

Solving

From the above discussion, it follows that d  d max when

  58.2  Finally, using Eq. (1) d

or

64 sin (2  58.229) 9.81

d max  5.84 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.116* A nozzle at A discharges water with an initial velocity of 36 ft/s at an angle  with the horizontal. Determine (a) the distance d to the farthest point B on the roof that the water can reach, (b) the corresponding angle  . Check that the stream will clear the edge of the roof.

SOLUTION Horizontal motion:

x   v0 cos   t

Vertical motion:

y  y0   v0 sin   t 

y  y0  x tan  

Let

u  x tan 

x2 

du

 xmax 

2



2 2  36   

32.2

(a) Maximum distance:



y  y0  u 



g x2  u 2 2v02



   0.

d x2 du

   2v

d x2

v0  36 ft/s,

1 2 1 gx 2 gt  y0  x tan   2 2  v0 cos  2

2v02  u  y0  y   u 2 g

The maximum value of x 2 is required:

Data:

x v0 cos 

gx 2 1  tan 2  2v02

so that

Solving for x 2 :

t

or

2 0

g

y0  3.6 ft,

 2u  0

or

yB  18 ft

u

v02 g

2 36   u

32.2

 40.2484  3.6  18   40.2484 2  460.78 ft 2 d  xmax  13.5

 40.2484 ft

xmax  21.466 ft

d  7.97 ft 

PROBLEM 11.116* (Continued) (b) Angle  .

tan  

xmax tan  u 40.2484    1.875 xmax xmax 21.466

  61.9 

  61.93 Check the edge.

y  y0  x tan  

gx 2 2  v0 cos  

y  3.6  13.5 1.875  

2

 32.2 13.5 2 2 2 36 cos 61.93

y  18.69 ft 

Since y  18 ft, the stream clears the edge.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.117 The velocities of skiers A and B are as shown. Determine the velocity of A with respect to B.

SOLUTION We have

vA  vB  vA/B

The graphical representation of this equation is then as shown. Then

v 2A/B  302  452  2(30)(45) cos 15

or

vA /B  17.80450 ft/s

and or

30 17.80450  sin  sin 15

  25.8554°   25  50.8554°

vA/B  17.8 ft/s

Alternative solution.

vA /B  vA  vB  30 cos 10 i  30 sin 10 j  (45 cos 25 i  45 sin 25 j)  11.2396i  13.8084 j  5.05 m/s  17.8 ft/s

50.9°

50.9° 

PROBLEM 11.118 The three blocks shown move with constant velocities. Find the velocity of each block, knowing that the relative velocity of A with respect to C is 300 mm/s upward and that the relative velocity of B with respect to A is 200 mm/s downward.

SOLUTION From the diagram Cable 1:

y A  yD  constant

Then

v A  vD  0

Cable 2:

(1)

( yB  yD )  ( yC  yD )  constant vB  vC  2vD  0

Then

(2)

Combining Eqs. (1) and (2) to eliminate vD ,

2v A  vB  vC  0

(3)

Now

v A/C  v A  vC  300 mm/s

(4)

and

vB/A  vB  v A  200 mm/s

(5)

(3)  (4)  (5) 

Then

(2v A  vB  vC )  (v A  vC )  (vB  v A )  (300)  (200) v A  125 mm/s 

or and using Eq. (5)

vB  (125)  200 v B  75 mm/s 

or Eq. (4) or

125  vC  300 vC  175 mm/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.119 Three seconds after automobile B passes through the intersection shown, automobile A passes through the same intersection. Knowing that the speed of each automobile is constant, determine (a) the relative velocity of B with respect to A, (b) the change in position of B with respect to A during a 4-s interval, (c) the distance between the two automobiles 2 s after A has passed through the intersection.

SOLUTION

v A  45 mi/h  66 ft/s vB  30 mi/h  44 ft/s Law of cosines vB2/A  662  442  2(66)(44) cos110 vB/A  90.99 ft/s

vB  v A  vB/A

Law of sines sin  sin110  66 90.99

  42.97

  90    90  42.97  47.03 v B/A  91.0 ft/s

(a)

Relative velocity:

(b)

Change in position for t  4 s. rB/A  vB/A t  (91.0 ft/s)(4 s)

(c)

Distance between autos 2 seconds after auto A has passed intersection. Auto A travels for 2 s.

v A  66 ft/s

20°

rA  v At  (66 ft/s)(2 s)  132 ft rA  132 ft

20°

rB/A  364 ft

47.0° 

47.0° 

PROBLEM 11.119 (Continued)

Auto B

v B  44 ft/s rB  v B t  (44 ft/s)(5 s)  220 ft

rB  rA  rB/A Law of cosines rB2/A  (132) 2  (220) 2  2(132)(220) cos110 rB/A  292.7 ft

Distance between autos  293 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.120 Shore-based radar indicates that a ferry leaves its slip with a 70°, while instruments aboard the velocity v  18 km/h ferry indicate a speed of 18.4 km/h and a heading of 30° west of south relative to the river. Determine the velocity of the river.

SOLUTION

We have

v F  v R  v F/R

or

v F  v F/R  v R

The graphical representation of the second equation is then as shown. We have

vR2  182  18.42  2(18)(18.4) cos 10

or

vR  3.1974 km/h

and or

18 3.1974  sin  sin 10

  77.84

Noting that

v R  3.20 km/h Alternatively one could use vector algebra.

17.8° 

PROBLEM 11.121 Airplanes A and B are flying at the same altitude and are tracking the eye of hurricane C. The relative velocity of C 75°, and the relative with respect to A is vC/A  350 km/h 40°. velocity of C with respect to B is vC/B  400 km/h Determine (a) the relative velocity of B with respect to A, (b) the velocity of A if ground-based radar indicates that the hurricane is moving at a speed of 30 km/h due north, (c) the change in position of C with respect to B during a 15-min interval.

SOLUTION (a)

We have

vC  v A  vC/A

and

vC  v B  vC/B

Then

v A  vC/A  v B  vC/B

or

v B  v A  v C /A  v C /B

Now

v B  v A  v B /A

so that

v B/A  v C/A  v C/B

or

v C/A  v C/B  v B/A

The graphical representation of the last equation is then as shown. We have

vB2/A  3502  4002  2(350)(400) cos 65

or

vB/A  405.175 km/h

and or

400 405.175  sin  sin 65

  63.474 75    11.526

v B/A  405 km/h

11.53° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.121 (Continued)

(b)

We have

vC  v A  vC/A

or

v A  (30 km/h) j  (350 km/h)( cos 75i  sin 75 j ) v A  (90.587 km/h)i  (368.07 km/h) j v A  379 km/h

or (c)

76.17° 

Noting that the velocities of B and C are constant, we have rB  (rB )0  v B t

Now

rC  (rC )0  vC t

rC/B  rC  rB  [(rC )0  (rB )0 ]  ( vC  v B )t

 [(rC )0  (rB )0 ]  vC/B t Then

rC/B  (rC/B )t2  (rC/B )t1  vC/B (t2  t1 )  vC/B t

For t  15 min:

1  rC/B  (400 km/h)  h   100 km 4 

rC/B  100 km

40° 

PROBLEM 11.122 Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is 120 km/h and the direction of the relative velocity vector (heading) is 70° east of north. Instruments on the ground indicate that the velocity of the airplane (ground speed) is 110 km/h and the direction of flight (course) is 60° east of north. Determine the wind speed and direction.

SOLUTION Let i and j be unit vectors in directions east and north respectively. Velocity of plane relative to air. v P/ A  120 km/h  cos 20 i  sin 20 j

Velocity of plane.

v P  110 km/h  cos30 i  sin 30j But

v P  v A  v P/ A

Velocity of air.

v A  v P  v P/ A

v A  110cos30  120cos 20  i  110sin 30  120sin 20  j    17.50 km/h  i  13.96 km/h  j vA 

17.50 2  13.96 2

tan  

17.50 13.96

 22.4 km/h

  51.4 v A  22.4 km/h at 51.4 west of north 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

Problem 11.123 Knowing that at the instant shown block B has a velocity of 2 ft/s to the right and an acceleration of 3 ft/s2 to the left, determine (a) the velocity of block A, (b) the acceleration of block A.

SOLUTION Given:

vB  2 ft/s, aB  3 ft/s

Length of Cable

L  y A  2 xB  constants

Differentiate Length Equation

0  vA  2vB

(1)

Differentiate again

0  a A  2aB

(2)

(e) Rearrange (1)

vA  2vB

 2  2  ft/s

vA  4 ft/s (f) Rearrange (2)

v A  4 ft/s  

a A  2aB  2  3 ft/s 2 a A  6 ft/s 2

a A  6 ft/s 2  

PROBLEM 11.124 Knowing that at the instant shown block A has a velocity of 8 in./s and an acceleration of 6 in./s2 both directed down the incline, determine (a) the velocity of block B, (b) the acceleration of block B.

SOLUTION From the diagram

2 x A  xB/A  constant

Then

2v A  vB/A  0 | vB/A |  16 in./s

or

2a A  aB/A  0

and

| aB/A |  12 in./s2

or

Note that v B/A and a B/A must be parallel to the top surface of block A. (a)

We have

v B  v A  v B/A

The graphical representation of this equation is then as shown. Note that because A is moving downward, B must be moving upward relative to A. We have

vB2  82  162  2(8)(16) cos 15

or

vB  8.5278 in./s

and or

8 8.5278  sin  sin 15

  14.05 v B  8.53 in./s

(b)

54.1° 

The same technique that was used to determine v B can be used to determine a B . An alternative method is as follows. We have

a B  a A  a B/A  (6i )  12( cos 15i  sin 15 j)*  (5.5911 in./s 2 )i  (3.1058 in./s 2 ) j

or

a B  6.40 in./s2

54.1° 

* Note the orientation of the coordinate axes on the sketch of the system.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.125 A boat is moving to the right with a constant deceleration of 0.3 m/s2 when a boy standing on the deck D throws a ball with an initial velocity relative to the deck which is vertical. The ball rises to a maximum height of 8 m above the release point and the boy must step forward a distance d to catch it at the same height as the release point. Determine (a) the distance d, (b) the relative velocity of the ball with respect to the deck when the ball is caught.

SOLUTION Horizontal motion of the ball:

vx  (vx )0 ,

Vertical motion of the ball:

v y  (v y )0  gt yB  (v y )0 t 

At maximum height,

xball  (vx )0 t

1 2 gt , (v y ) 2  (v y )02  2 gy 2 vy  0 and y  ymax

(v y )2  2 gymax  (2)(9.81)(8)  156.96 m 2 /s 2 (v y )0  12.528 m/s At time of catch,

tcatch  2.554 s

or Motion of the deck:

1 (9.81)t 2 2 and v y  12.528 m/s

y  0  12.528 

v x  (v x ) 0  a D t ,

xdeck  (vx )0 t 

1 aDt 2 2

Motion of the ball relative to the deck: (vB/D ) x  (vx )0  [(vx )0  aDt ]  aDt 1 1   xB/D  (vx )0 t  (vx )0 t  aDt 2    aDt 2 2 2   (vB/D ) y  (v y )0  gt , yB/D  yB

(a) (b)

At time of catch,

1 d  xD/B   ( 0.3)(2.554) 2 2 (vB/D ) x  ( 0.3)(2.554)   0.766 m/s

(vB/D ) y  12.528 m/s

d  0.979 m 

or 0.766 m/s v B/D  12.55 m/s

86.5 

PROBLEM 11.126 The assembly of rod A and wedge B starts from rest and moves to the right with a constant acceleration of 2 mm/s2. Determine (a) the acceleration of wedge C, (b) the velocity of wedge C when t  10 s.

SOLUTION (a)

We have

aC  a B  aC/B

The graphical representation of this equation is then as shown. First note

  180  (20  105)  55

Then

aC 2  sin 20 sin 55 aC  0.83506 mm/s 2

(b)

aC  0.835 mm/s2

75° 

v C  8.35 mm/s

75° 

For uniformly accelerated motion

vC  0  aC t At t  10 s:

vC  (0.83506 mm/s 2 )(10 s)  8.3506 mm/s

or

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.127 Determine the required velocity of the belt B if the relative velocity with which the sand hits belt B is to be (a) vertical, (b) as small as possible.

SOLUTION A grain of sand will undergo projectile motion.

vsx  vsx  constant  5 ft/s 0

vs y  2 gh  (2)(32.2 ft/s 2 )(3 ft)  13.90 ft/s 

y-direction.

v S/B  v S  v B

Relative velocity. (a)

(1)

If vS/B is vertical,

vS/B j  5i  13.9 j  (vB cos 15i  vB sin 15 j)  5i  13.9 j  vB cos 15i  vB sin 15 j Equate components.

i : 0  5  vB cos 15

vB 

5  5.176 ft/s cos 15 v B  5.18 ft/s

(b)

15° 

vS/C is as small as possible, so make vS/B  to vB into (1). vS/B sin 15i  vS/B cos 15 j   5i  13.9 j  vB cos 15i  vB sin 15 j Equate components and transpose terms.

(sin 15) vS/B  (cos 15) vB  5 (cos 15) vS/B  (sin 15) vB  13.90 Solving,

vS/B  14.72 ft/s

vB  1.232 ft/s v B  1.232 ft/s

15° 

PROBLEM 11.128 Conveyor belt A, which forms a 20° angle with the horizontal, moves at a constant speed of 4 ft/s and is used to load an airplane. Knowing that a worker tosses duffel bag B with an initial velocity of 2.5 ft/s at an angle of 30° with the horizontal, determine the velocity of the bag relative to the belt as it lands on the belt.

SOLUTION First determine the velocity of the bag as it lands on the belt. Now

[(vB ) x ]0  (vB )0 cos 30  (2.5 ft/s) cos 30 [(vB ) y ]0  (vB )0 sin 30  (2.5 ft/s)sin 30 Horizontal motion. (Uniform)

x  0  [(vB ) x ]0 t

(vB ) x  [(vB ) x ]0

 (2.5 cos 30)t

 2.5 cos 30

Vertical motion. (Uniformly accelerated motion) y  y0  [(vB ) y ]0 t 

1 2 gt 2

 1.5  (2.5 sin 30) t 

(vB ) y  [(vB ) y ]0  gt

1 2 gt 2

 2.5 sin 30  gt

The equation of the line collinear with the top surface of the belt is

y  x tan 20 Thus, when the bag reaches the belt 1.5  (2.5 sin 30) t 

1 2 gt  [(2.5 cos 30) t ]tan 20 2

or

1 (32.2) t 2  2.5(cos 30 tan 20  sin 30) t  1.5  0 2

or

16.1t 2  0.46198t  1.5  0

Solving

t  0.31992 s and t  0.29122 s (Reject)

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.128 (Continued)

The velocity v B of the bag as it lands on the belt is then

v B  (2.5 cos 30)i  [2.5 sin 30  32.2(0.319 92)]j  (2.1651 ft/s)i  (9.0514 ft/s) j Finally or

v B  v A  v B/A

v B/A  (2.1651i  9.0514 j)  4(cos 20i  sin 20 j)  (1.59367 ft/s)i  (10.4195 ft/s) j

or

v B/A  10.54 ft/s

81.3° 

PROBLEM 11.129 During a rainstorm the paths of the raindrops appear to form an angle of 30° with the vertical and to be directed to the left when observed from a side window of a train moving at a speed of 15 km/h. A short time later, after the speed of the train has increased to 24 km/h, the angle between the vertical and the paths of the drops appears to be 45°. If the train were stopped, at what angle and with what velocity would the drops be observed to fall?

SOLUTION vrain  vtrain  vrain/train Case :

vT  15 km/h

;

vR / T

30°

Case :

vT  24 km/h

;

vR / T

45°

Case :

(vR ) y tan 30  15  (vR ) x

(1)

Case :

(vR ) y tan 45  24  (vR ) x

(2)

Substract (1) from (2)

(vR ) y (tan 45  tan 30)  9 (vR ) y  21.294 km/h

Eq. (2):

21.294 tan 45  25  (vR ) x (vR ) x  2.706 km/h

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.129 (Continued)

3.706 21.294   7.24 21.294  21.47 km/h  5.96 m/s vR  cos 7.24

tan  

vR  5.96 m/s

82.8° 

Alternate solution Alternate, vector equation

v R  vT  v R / T

For first case,

v R  15i  vR /T 1 ( sin 30i  cos 30 j)

For second case,

v R  24i  vR /T  2 ( sin 45i  cos 45 j)

Set equal 15i  vR /T 1 ( sin 30i  cos 30 j)  24i  vR /T  2 ( sin 45i  cos 45 j) Separate into components:

i:

15  vR /T 1 sin 30  24  vR /T  2 sin 45 vR /T 1 sin 30  vR /T  2 sin 45  9

(3)

vR /T 1 cos30  vR /T  2 cos 45

j:

vR /T 1 cos 30  vR /T  2 cos 45  0

(4)

Solving Eqs. (3) and (4) simultaneously, vR /T 1  24.5885 km/h

vR /T  2  30.1146 km/h

Substitute v R /T  2 back into equation for v R .

v R  24i  30.1146( sin 45i  cos 45 j) v R  2.71i  21.29 j  21.29    82.7585  2.71 

  tan 1 

v R  21.4654 km/hr  5.96 m/s v R  5.96 m/s

82.8° 

PROBLEM 11.130 Instruments in airplane A indicate that with respect to the air the plane is headed 30o north of east with an airspeed of 300 mi/h. At the same time radar on ship B indicates that the relative velocity of the plane with respect to the ship is 280 mi/h in the direction 33o north of east. Knowing that the ship is steaming due south at 12 mi/h, determine (a) the velocity of the airplane, (b) the wind speed and direction.

SOLUTION Let the x-axis be directed east, and the y-axis be directed north. Airspeed:

30  300  cos 30i  sin 30 j mi/h

v AW  300 mi/h /

v A/B  280  cos 33i  sin 33 j  mi/h

Plane relative to ship:

v B  12 mi/h

Ship:

 12 j

(a) Velocity of airplane. v A  v B  v A/B  12 j  280  cos 33i  sin 33 j 

 234.83i  140.50 j

v A  274 mi/h

30.9 

vW  26.7 mi/h

20.8 

(b) Wind velocity.

vW  v A  vW / A  v A  v A/W

 234.83i  140.50 j  300  cos30i  sin 30j  24.98i  9.50 j

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.131 When a small boat travels north at 5 km/h, a flag mounted on its stern forms an angle   50 with the centerline of the boat as shown. A short time later, when the boat travels east at 20 km/h, angle  is again 50°. Determine the speed and the direction of the wind.

SOLUTION We have

vW  v B  vW/B

Using this equation, the two cases are then graphically represented as shown.

With vW now defined, the above diagram is redrawn for the two cases for clarity.

Noting that

  180  (50  90   )  40   We have

vW 5  sin 50 sin (40   )

  180  (50   )  130  

vW 20  sin 50 sin (130   )

PROBLEM 11.131 (Continued)

Therefore or or

5 20  sin (40   ) sin (130   ) sin 130 cos   cos 130 sin   4(sin 40 cos   cos 40 sin  ) tan  

sin 130  4 sin 40 cos 130  4 cos 40

or

  25.964

Then

vW 

5 sin 50  15.79 km/h sin (40  25.964)

vW  15.79 km/h

26.0° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.132 As part of a department store display, a model train D runs on a slight incline between the store’s up and down escalators. When the train and shoppers pass Point A, the train appears to a shopper on the up escalator B to move downward at an angle of 22° with the horizontal, and to a shopper on the down escalator C to move upward at an angle of 23° with the horizontal and to travel to the left. Knowing that the speed of the escalators is 3 ft/s, determine the speed and the direction of the train.

SOLUTION v D  v B  v D/B

We have

v D  vC  v D/C The graphical representations of these equations are then as shown. Then

vD 3  sin 8 sin (22   )

Equating the expressions for

vD 3  sin 7 sin (23   )

vD 3

sin 8 sin 7  sin (22   ) sin (23   ) or

or or Then

sin 8 (sin 23 cos   cos 23 sin  )  sin 7 (sin 22 cos   cos 22 sin  ) tan  

sin 8 sin 23  sin 7 sin 22 sin 8 cos 23  sin 7 cos 22

  2.0728 vD 

3 sin 8  1.024 ft/s sin (22  2.0728) v D  1.024 ft/s

2.07° 

PROBLEM 11.132 (Continued)

Alternate solution using components.

v B  (3 ft/s)

30  (2.5981 ft/s)i  (1.5 ft/s) j

vC  (3 ft/s)

30  (2.5981 ft/s)i  (1.5 ft/s) j

v D/B  u1

22  (u1 cos 22)i  (u1 sin 22) j

v D/C  u2

23  (u2 cos 23)i  (u2 sin 23) j

v D  vD

  (vD cos  )i  (vD sin  ) j

v D  v B  v D/B  vC  v D/C

2.5981i  1.5 j  (u1 cos 22)i  (u1 sin 22) j  2.5981i  1.5 j  (u2 cos 23)i  (u2 sin 23) j Separate into components, transpose, and change signs.

u1 cos 22  u2 cos 23  0 u1 sin 22  u1 sin 23  3 Solving for u1 and u2 ,

u1  3.9054 ft/s

u2  3.9337 ft/s

v D  2.5981i  1.5 j  (3.9054 cos 22)i  (3.9054 sin 22) j

 (1.0229 ft/s)i  (0.0370 ft/s) j or

v D  2.5981i  1.5 j  (3.9337 cos 23)i  (3.9337 sin 23) j  (1.0229 ft/s)i  (0.0370 ft/s) j v D  1.024 ft/s

2.07° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.133 Determine the smallest radius that should be used for a highway if the normal component of the acceleration of a car traveling at 72 km/h is not to exceed 0.8 m/s 2 .

SOLUTION an 

v2

an  0.8 m/s 2



v  72 km/h  20 m/s

0.8 m/s 2 

(20 m/s)2



  500 m  

PROBLEM 11.134 Determine the maximum speed that the cars of the roller-coaster can reach along the circular portion AB of the track if  is 25 m and the normal component of their acceleration cannot exceed 3 g.

SOLUTION We have

an 

v2



Then

(vmax )2AB  (3  9.81 m/s2 )(25 m)

or

(vmax ) AB  27.124 m/s

or

(vmax ) AB  97.6 km/h 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

Problem 11.135 Human centrifuges are often used to simulate different acceleration levels for pilots and astronauts. Space shuttle pilots typically face inwards towards the center of the gondola in order to experience a simulated 3g forward acceleration. Knowing that the astronaut sits 5 m from the axis of rotation and experiences 3 g’s inward, determine her velocity.

SOLUTION Given:

an  3g ,   5 m an 

Rearrange

v2



v   an v   3g

Substitute in known values

v  5*3*9.81 m/s vA  12.13 m/s 

PROBLEM 11.136 The diameter of the eye of a stationary hurricane is 20 mi and the maximum wind speed is 100 mi/h at the eye wall, r  10 mi. Assuming that the wind speed is constant for constant r and decreases uniformly with increasing r to 40 mi/h at r  110 mi, determine the magnitude of the acceleration of the air at (a) r  10 mi, (b) r  60 mi, (c) r  110 mi.

SOLUTION (a)

r  10 mi  52800 ft,

a (b)

v  100 mi/h  146.67 ft/s

146.67  v2  r 52800

r  60 mi  316800 ft, v  100 

2

40  100  60  10   70 mi/h  102.67 ft/s 110  10

102.67  v2 a  r 316800 (c)

a  0.407 ft/s2 

2

a  0.0333 ft/s2 

r  110 mi  580800 ft, v  40 mi/h  58.67 ft/s

 58.67  v2 a  r 580800

2

a  0.00593 ft/s2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.137 The peripheral speed of the tooth of a 10-in.-diameter circular saw blade is 150 ft/s when the power to the saw is turned off. The speed of the tooth decreases at a constant rate, and the blade comes to rest in 9 s. Determine the time at which the total acceleration of the tooth is 130 ft/s2.

SOLUTION v  v0  at t

For uniformly decelerated motion: At t  9 s,

0  150  at  9  ,

at  16.667 ft/s 2

a 2  at2  an2

Total acceleration: 1/2

an   a 2  at2 

Normal acceleration:

or

an 

1/2

2 2  130    16.667    

v2



,

where

 

 5 v 2   an    128.93  53.72 ft 2 /s 2 ,  12  Time:

t 

v  v0 7.329  150  16.667 at

 128.93 ft/s 2

1 5 diameter  ft 2 12 v  7.329 ft/s

t  8.56 s 

PROBLEM 11.138 A robot arm moves so that P travels in a circle about Point B, which is not moving. Knowing that P starts from rest, and its speed increases at a constant rate of 10 mm/s2, determine (a) the magnitude of the acceleration when t  4 s, (b) the time for the magnitude of the acceleration to be 80 mm/s2.

SOLUTION Tangential acceleration:

at  10 mm/s 2 v  at t

Speed:

v2

an 

where

  0.8 m  800 mm

(a)

When t  4 s



v  (10)(4)  40 mm/s an 

Acceleration:





at2t 2

Normal acceleration:

(40) 2  2 mm/s 2 800

a  at2  an2  (10)2  (2)2 a  10.20 mm/s 2 

(b)

Time when a  80 mm/s 2 a 2  an2  at2 2

 (10)2 t 2  2 (80)     10 800   2

t 4  403200 s 4 t  25.2 s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.139 A monorail train starts from rest on a curve of radius 400 m and accelerates at the constant rate at . If the maximum total acceleration of the train must not exceed 1.5 m/s2 , determine (a) the shortest distance in which the train can reach a speed of 72 km/h, (b) the corresponding constant rate of acceleration at .

SOLUTION When v  72 km/h  20 m/s and   400 m,

an 

v2





(20)2  1.000 m/s 2 400

a  an2  at2

But

at  a 2  an2  (1.5)2  (1.000)2  1.11803 m/s2 Since the train is accelerating, reject the negative value. (a)

Distance to reach the speed. v0  0 Let

x0  0

v12  v02  2at ( x1  x0 )  2at x1 x1  (b)



v12 (20)2  2at (2)(1.11803)

x1  178.9 m 

Corresponding tangential acceleration.

at  1.118 m/s2 

PROBLEM 11.140 A motorist starts from rest at Point A on a circular entrance ramp when t  0, increases the speed of her automobile at a constant rate and enters the highway at Point B. Knowing that her speed continues to increase at the same rate until it reaches 100 km/h at Point C, determine (a) the speed at Point B, (b) the magnitude of the total acceleration when t  20 s.

SOLUTION v0  0

Speeds:

v1  100 km/h  27.78 m/s s

Distance:

 2

(150)  100  335.6 m

v12  v02  2at s

Tangential component of acceleration:

at  At Point B,

v12  v02 (27.78)2  0   1.1495 m/s 2 2s (2)(335.6)

vB2  v02  2at sB

where

sB 

 2

(150)  235.6 m

vB2  0  (2)(1.1495)(235.6)  541.69 m2 /s 2 vB  23.27 m/s (a)

At t  20 s,

vB  83.8 km/h 

v  v0  at t  0  (1.1495)(20)  22.99 m/s

  150 m

Since v  vB , the car is still on the curve.

(b)

Normal component of acceleration:

an 

Magnitude of total acceleration:

|a| 

v2





(22.99)2  3.524 m/s 2 150

at2  an2 

(1.1495)2  (3.524)2

| a |  3.71 m/s2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.141 Racecar A is traveling on a straight portion of the track while racecar B is traveling on a circular portion of the track. At the instant shown, the speed of A is increasing at the rate of 10 m/s2, and the speed of B is decreasing at the rate of 6 m/s2. For the position shown, determine (a) the velocity of B relative to A, (b) the acceleration of B relative to A.

SOLUTION v A  240 km/h  66.67 m/s

Speeds:

vB  200 km/h  55.56 m/s vA  66.67 m/s

Velocities:

vB  55.56 m/s (a)

50°

vB/A  vB  vA

Relative velocity:

vB/A  (55.56 cos 50)   55.56 sin 50   66.67

 30.96   42.56  52.63 m/s

53.96°

vB /A  189.5 km/h Tangential accelerations:

(aA )t  10 m/s2 (aB )t  6 m/s 2 an 

Normal accelerations:



(   )

Car B:

(   300 m)

Acceleration of B relative to A:

50°

v2

Car A:

(aB )n  (b)

54.0° 

(aA ) n  0

(55.56)2  10.288 300

(aB )n  10.288 m/s 2

40°

aB/A  aB  aA a B/A  (a B )t  (a B )n  (a A )t  (a A )n 6

50  10.288

40  10   0

 (6cos 50  10.288cos 40  10)  (6sin 50  10.288sin 40)  21.738   2.017

a B/A  21.8 m/s 2

5.3° 

PROBLEM 11.142 At a given instant in an airplane race, airplane A is flying horizontally in a straight line, and its speed is being increased at the rate of 8 m/s 2 . Airplane B is flying at the same altitude as airplane A and, as it rounds a pylon, is following a circular path of 300-m radius. Knowing that at the given instant the speed of B is being decreased at the rate of 3 m/s2 , determine, for the positions shown, (a) the velocity of B relative to A, (b) the acceleration of B relative to A.

SOLUTION First note

v A  450 km/h vB  540 km/h  150 m/s

(a)

v B  v A  v B/A

We have

The graphical representation of this equation is then as shown. We have

vB2/A  4502  5402  2(450)(540) cos 60° vB/A  501.10 km/h

and

540 501.10  sin  sin 60

  68.9 v B/A  501 km/h (b)

First note Now

a A  8 m/s2 (aB ) n 

v 2B

B



(a B )t  3 m/s 2

Then

60

(150 m/s) 2 300 m

(a B )n  75 m/s2 

68.9 

30 

a B  (a B )t + (a B ) n  3( cos 60 i  sin 60 j) + 75(cos 30 i  sin 30 j) = (66.452 m/s 2 )i  (34.902 m/s 2 ) j

Finally

a B  a A  a B/ A a B/A  (66.452i  34.902 j)  (8i )

 (74.452 m/s 2 )i  (34.902 m/s 2 ) j

a B/A  82.2 m/s2

25.1 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.143 A race car enters the circular portion of a track that has a radius of 70 m. When the car enters the curve at point P, it is travelling with a speed of 120 km/h that is increasing at 5 m/s2. Three seconds later, determine (a) the total acceleration of the car in xy components, (b) the linear velocity of the car in xy components.

SOLUTION Speeds:

vto  120 km/h  33.33 m/s

vt  vto  at t vt  33.33  5(3)  48.33 m/s

Tangential Acceleration:

1 s  so  vto t  at t 2 2 1 s  33.33(3)  (5)(3)2  122.5 m 2 s  R s 122.5  1.75 rad   R 70   100.3 at  5 m/s 2

Normal Acceleration:

an 

Distance:

Location:

an  (a)

Accelerations:

v2

 48.332  33.37 m/s 2 70

ax  5cos(10.3)  33.37sin(10.3) a x  1.047 m/s 2  a y  5 sin(10.3)  33.37 cos(10.3)

a y  33.726 m/s 2  (b)

Velocities:

vx  48.33cos(10.3) vx  47.55 m/s  v y  48.33sin(10.3) v y  8.64 m/s 

PROBLEM 11.144 An airplane flying at a constant speed of 240 m/s makes a banked horizontal turn. What is the minimum allowable radius of the turn if the structural specifications require that the acceleration of the airplane shall never exceed 4g?

SOLUTION Given:

v  240 m/s, amax  4 g

Acceleration:

an 

Rearrange:



v2 an

Substitute in known values:



2402 m 4*9.81

v2



  1467.9 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.145 A golfer hits a golf ball from Point A with an initial velocity of 50 m/s at an angle of 25° with the horizontal. Determine the radius of curvature of the trajectory described by the ball (a) at Point A, (b) at the highest point of the trajectory.

SOLUTION (a)

We have

or

(a A ) n 

A 

v A2

A (50 m/s) 2 (9.81 m/s 2 ) cos 25

 A  281 m 

or (b)

We have

( aB ) n 

vB2

B

where Point B is the highest point of the trajectory, so that

vB  (v A ) x  v A cos 25 Then or

B 

[(50 m/s) cos 25°]2 9.81 m/s 2

 B  209 m 

PROBLEM 11.146 Three children are throwing snowballs at each other. Child A throws a snowball with a horizontal velocity v0. If the snowball just passes over the head of child B and hits child C, determine the radius of curvature of the trajectory described by the snowball (a) at Point B, (b) at Point C.

SOLUTION The motion is projectile motion. Place the origin at Point A. Horizontal motion:

v x  v0

x  v0t

Vertical motion:

y0  0,

(v y )  0

v y   gt

1 y   gt 2 2

2h , g

t

where h is the vertical distance fallen.

| v y|  2 gh v 2  vx2  v 2y  v02  2 gh

Speed: Direction of velocity.

cos  

v0 v

Direction of normal acceleration.

an  g cos  

Radius of curvature: At Point B,



gv0 v 2  v 

v3 gv0

hB  1 m; xB  7 m

tB 

(2)(1 m)  0.45152 s 9.81 m/s 2

xB  v0 t B

v0 

xB 7m   15.504 m/s t B 0.45152 s

vB2  (15.504) 2  (2)(9.81)(1)  259.97 m 2 /s 2

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.146 (Continued) (a)

Radius of curvature at Point B.

B  At Point C

(259.97 m 2 /s 2 )3/ 2 (9.81 m/s 2 )(15.504 m/s)

 B  27.6 m 

hC  1 m  2 m  3 m

vC2  (15.504)2  (2)(9.81)(3)  299.23 m 2 /s2 (b)

Radius of curvature at Point C.

C 

(299.23 m 2 /s 2 )3/2 (9.81 m/s 2 )(15.504 m/s)

C  34.0 m 

PROBLEM 11.147 Coal is discharged from the tailgate A of a dump truck with an initial velocity v A  2 m/s 50 . Determine the radius of curvature of the trajectory described by the coal (a) at Point A, (b) at the point of the trajectory 1 m below Point A.

SOLUTION  9.81 m/s 2

aA  g

(a) At Point A.

Sketch tangential and normal components of acceleration at A. (a A ) n  g cos 50

A 

v A2 (2)2  (a A ) n 9.81cos50

 A  0.634 m 

(b) At Point B, 1 meter below Point A. Horizontal motion: (vB ) x  (v A ) x  2 cos 50  1.286 m/s

(vB )2y  (vA )2y  2a y ( yB  y A )

Vertical motion:

 (2 cos 40) 2  (2)(9.81)(1)  21.97 m 2 /s 2

(vB ) y  4.687 m/s tan  

(vB ) y (vB ) x



4.687 , 1.286

or

  74.6

aB  g cos 74.6

B 



(vB )2x  (vB )2y vB 2  (a B ) n g cos 74.6 (1.286)2  21.97 9.81cos 74.6

 B  9.07 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.148 From measurements of a photograph, it has been found that as the stream of water shown left the nozzle at A, it had a radius of curvature of 25 m. Determine (a) the initial velocity vA of the stream, (b) the radius of curvature of the stream as it reaches its maximum height at B.

SOLUTION (a)

We have

(a A ) n 

v A2

A

or

4  v A2   (9.81 m/s 2 )  (25 m) 5 

or

v A  14.0071 m/s vA  14.01 m/s

(b)

We have Where

Then or

( aB ) n 

vB2

B

vB  (v A ) x 

B

36.9° 

4 vA 5

 4  14.0071 m/s   5

2

9.81 m/s2

 B  12.80 m 

PROBLEM 11.149 A child throws a ball from point A with an initial velocity v0 at an angle of 3 with the horizontal. Knowing that the ball hits a wall at point B, determine (a) the magnitude of the initial velocity, (b) the minimum radius of curvature of the trajectory.

SOLUTION Horizontal motion.

vx  v0 cos 

x  v0 t cos 

Vertical motion.

v y  v0 sin   gt y  y0  v0 t sin  

y  y0  x tan  

Eliminate t.

1 2 gt 2

gx 2 2 v02 cos 2 

(1)

Solving (1) for v0 and applying result at point B v0 

gx 2  2  y0  x tan   y  cos 2 

 2 1.5  6 tan 3  0.97   cos2 3  v0  14.48 m/s 

(a)

Magnitude of initial velocity.

(b)

Minimum radius of curvature of trajectory.

an  g 

 9.81 6 2

v2





v2 v2  an g cos

(2)

where  is the slope angle of the trajectory. The minimum value of  occurs at the highest point of the trajectory where cos   1 and v  vx  v0 cos  Then 2 v02 cos 2  14.48 cos 3   g 9.81 2

 min

 min  21.3 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.150 A projectile is fired from Point A with an initial velocity v 0 . (a) Show that the radius of curvature of the trajectory of the projectile reaches its minimum value at the highest Point B of the trajectory. (b) Denoting by  the angle formed by the trajectory and the horizontal at a given Point C, show that the radius of curvature of the trajectory at C is   min /cos3  .

SOLUTION For the arbitrary Point C, we have

(aC ) n  or

C 

vC2

C vC2 g cos 

Noting that the horizontal motion is uniform, we have

(v A ) x  (vC ) x where Then

(v A ) x  v0 cos  v0 cos   vC cos  cos  v0 cos 

or

vC 

so that

1 C  g cos 

(a)

2

 cos   v 2 cos2  v0   0  g cos3   cos  

In the expression for C , v0 ,  , and g are constants, so that C is minimum where cos  is maximum. By observation, this occurs at Point B where   0.

 min   B  (b)

(vC ) x  vC cos 

C 

1 cos3 

C 

 min cos3 

v02 cos 2  g

 v02 cos 2   g 

Q.E.D.



Q.E.D.



  

PROBLEM 11.151* Determine the radius of curvature of the path described by the particle of Problem 11.95 when t  0.

PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r  (Rt cos nt)i  ctj  (Rt sin nt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.)

SOLUTION We have

v

dr  R(cos n t  n t sin n t )i  cj  R(sin n t  n t cos n t )k dt

and

a

dv  R  n sin n t  n sin n t  n2t cos n t i dt









 R n cos n t  n cos n t  n2t sin n t k a  n R [(2 sin n t  n t cos n t )i  (2cos n t  n t sin n t ) k ]

or

v 2  R 2 (cos n t  n t sin n t ) 2  c 2  R 2 (sin n t  n t cos n t ) 2

Now





 R 2 1  n2 t 2  c 2





1/ 2

v   R 2 1  n2t 2  c 2   

Then

R 2n2t dv  1/ 2 dt  2 R 1  n2t 2  c 2   

and





2 2  dv   v a 2  at2  an2       dt   

Now At t  0:

  

2

dv 0 dt a  n R(2 k ) or a  2n R v2  R2  c2

Then, with we have or

dv  0, dt

a 2n R 

v2

 R2  c2





R2  c2  2n R

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.152* Determine the radius of curvature of the path described by the particle of Problem 11.96 when t  0, A  3, and B  1.

PROBLEM 11.96 The three-dimensional motion of a particle is defined by the position vector r  ( At cos t )i  ( A t 2  1) j  ( Bt sin t )k , where r and t are expressed in feet and seconds, respectively. Show that the curve described by the particle lies on the hyperboloid (y/A)2  (x/A)2  (z/B)2  1. For A  3 and B  1, determine (a) the magnitudes of the velocity and acceleration when t  0, (b) the smallest nonzero value of t for which the position vector and the velocity are perpendicular to each other.

SOLUTION With

A  3,

B 1

we have

r  (3t cos t )i  3 t 2  1 j  (t sin t )k

Now

v

and

a





 3t  dr j  (sin t  t cos t )k  3(cos t  t sin t )i   2  t  1  dt  

  2 dv  3( sin t  sin t  t cos t )i  3  t  1  t   dt t2  1   (cos t  cos t  t sin t )k   3(2sin t  t cos t )i  3

t t2  1

  j  

1 j (t  1)1/ 2 2

 (2 cos t  t sin t )k

v 2  9(cos t  t sin t )2  9

Then

t2  (sin t  t cos t )2 t 1 2

Expanding and simplifying yields

v 2  t 4  19t 2  1  8(cos2 t  t 4 sin 2 t )  8(t 3  t )sin 2t Then and

v  [t 4  19t 2  1  8(cos2 t  t 4 sin 2 t )  8(t 3  t )sin 2t ]1/ 2 dv 4t 3  38t  8(2cos t sin t  4t 3sin 2 t  2t 4sin t cos t )  8[(3t 2  1)sin 2t  2(t 3  t ) cos 2t ]  dt 2[t 4  19t 2  1  8(cos 2 t  t 4 sin 2 t )  8(t 3  t )sin 2t ]1/ 2

PROBLEM 11.152* (Continued) 2

a 

Now

at2



an2

2 2  dv   v         dt    

2

At t  0:

a  3 j  2k

or

a  13 ft/s2 dv 0 dt v 2  9 (ft/s) 2

Then, with

dv  0, dt

we have

a

or



v2

 9 ft 2 /s2 13 ft/s 2

  2.50 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.153 A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the acceleration of the satellite is equal to g ( R /r )2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s2 , determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular. Earth: (umean )orbit  107 Mm/h.

SOLUTION g  274 m/s 2 ,

For the sun, R

and

Given that an 

gR 2 v2 a  and that for a circular orbit n r r2

Eliminating an and solving for r,

r 

gR 2 v2

v  107  106 m/h  29.72  103 m/s

For the planet Earth, Then

1 1 D    (1.39  109 )  0.695  109 m 2 2

r 

(274)(0.695  109 ) 2  149.8  109 m (29.72  103 )2

r  149.8 Gm 

PROBLEM 11.154 A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the acceleration of the satellite is equal to g ( R/r )2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s2 , determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular. Saturn: (umean )orbit  34.7 Mm/h.

SOLUTION g  274 m/s 2

For the sun, R

and

Given that an 

1 1 D    (1.39  109 )  0.695  109 m 2 2

gR 2 v2 a  . and that for a circular orbit: n r r2 gR 2 v2

Eliminating an and solving for r,

r 

For the planet Saturn,

v  34.7  106 m/h  9.639  103 m/s

Then,

r 

(274)(0.695  109 ) 2  1.425  1012 m (9.639  103 )2

r  1425 Gm 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.155 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Venus: g  29.20 ft/s2 , R  3761 mi.

SOLUTION From Problems 11.153 and 11.154,

an 

gR 2 r2

For a circular orbit,

an 

v2 r

v R

Eliminating an and solving for v, For Venus,

g r

g  29.20 ft/s 2

R  3761 mi  19.858  106 ft. r  3761  100  3861 mi  20.386  106 ft Then,

v  19.858  106

29.20  23.766  103 ft/s 20.386  106 v  16200 mi/h 

PROBLEM 11.156 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Mars: g  12.17 ft/s2 , R  2102 mi.

SOLUTION From Problems 11.153 and 11.154,

an 

gR 2 r2

For a circular orbit,

an 

v2 r

v R

Eliminating an and solving for v, For Mars,

g r

g  12.17 ft/s2 R  2102 mi  11.0986  106 ft r  2102  100  2202 mi  11.6266  106 ft

Then,

v  11.0986  106

12.17  11.35  103 ft/s 6 11.6266  10 v  7740 mi/h 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.157 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Jupiter: g  75.35 ft/s 2 , R  44, 432 mi.

SOLUTION From Problems 11.153 and 11.154,

an 

gR 2 r2

For a circular orbit,

an 

v2 r

v R

Eliminating an and solving for v, For Jupiter,

g r

g  75.35 ft/s2 R  44432 mi  234.60  106 ft r  44432  100  44532 mi  235.13  106 ft

Then,

v  (234.60  106 )

75.35  132.8  103 ft/s 6 235.13  10 v  90600 mi/h 

PROBLEM 11.158 A satellite will travel indefinitely in a circular orbit around the earth if the normal component of its 2 acceleration is equal to g  R / r  , where g  9.81 m/s 2 , R = radius of the earth = 6370 km, and r = distance from the center of the earth to the satellite. Assuming that the orbit of the moon is a circle of radius 384  103 km , determine the speed of the moon relative to the earth.

SOLUTION gR 2 r2

Normal acceleration:

an 

Solve for v2:

v 2  ran 

Data:

g  9.81 m/s 2 ,

an 

and

v2





v2 r

gR 2 r

R  6370 km = 6.370  106 m

r  384  103 km = 384  106 m 2

v 

 9.81  6.370  106  384  10

v = 1.018 m/s

6

2

 1.0366  106 m 2 /s 2

v  3670 km/h 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.159 Knowing that the radius of the earth is 6370 km, determine the time of one orbit of the Hubble Space Telescope, knowing that the telescope travels in a circular orbit 590 km above the surface of the earth. (See information given in Problems 11.153–11.155.)

SOLUTION an  g

We have Then or

g

R2 r2

and an 

v2 r

R2 v2  r r2 vR

g r

where

r Rh

The circumference s of the circular orbit is equal to s  2 r

Assuming that the speed of the telescope is constant, we have

s  vtorbit Substituting for s and v

2 r  R or

torbit  

or

g torbit r

2 r 3/2 R g 2 [(6370  590) km]3/2 1h  2 1/2 3 6370 km [9.81  10 km/s ] 3600 s

torbit  1.606 h 

PROBLEM 11.160 Satellites A and B are traveling in the same plane in circular orbits around the earth at altitudes of 120 and 200 mi, respectively. If at t  0 the satellites are aligned as shown and knowing that the radius of the earth is R  3960 mi, determine when the satellites will next be radially aligned. (See information given in Problems 11.153–11.155.)

SOLUTION an  g

We have Then

g

R2 r2

R2 v2  r r2

and an  or

v2 r

vR

g r

r Rh

where The circumference s of a circular orbit is equal to

s  2 r

Assuming that the speeds of the satellites are constant, we have s  vT

Substituting for s and v 2 r  R or Now

T

g T r

2 r 3/ 2 2 ( R  h)3/ 2  R g R g

hB  hA  (T ) B  (T ) A

Next let time TC be the time at which the satellites are next radially aligned. Then, if in time TC satellite B completes N orbits, satellite A must complete ( N  1) orbits. Thus, TC  N (T ) B  ( N  1)(T ) A

or

 2 ( R  hB )3/ 2   2 ( R  hA )3/ 2  N   ( N  1)   g g  R   R 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.160 (Continued)

N

or



Then

( R  hA )3/ 2 ( R  hB )



1 3960  200 3960 120

3/ 2



 ( R  hA )

3/ 2

TC  N (T ) B  N

3/2





1 R  hB R  hA



3/ 2

1

 33.835 orbits 1

2 ( R  hB )3/2 R g

(3960  200) mi  1 h 2  33.835 3960 mi 32.2 ft/s2  1 mi 1/ 2 3600s 5280 ft 3/ 2





TC  51.2 h 

or Alternative solution

From above, we have (T ) B  (T ) A . Thus, when the satellites are next radially aligned, the angles  A and  B swept out by radial lines drawn to the satellites must differ by 2 . That is,

 A   B  2 For a circular orbit

s  r

From above

s  vt and v  R

Then



At time TC : or

R g ( R  hA )

3/ 2

TC 

TC  

R g R g s vt 1  g   R t  3/ 2 t  t  r r r  r  ( R  h)3/ 2 r R g ( R  hB )3/ 2

TC  2

2 R g  ( R  h1 )3/ 2  ( R  h1 )3/ 2  A B   2



1 mi (3960 mi) 32.2 ft/s 2  5280 ft

 

or

g r

1 (3960  120) mi3/ 2

1 



1/ 2

1

(3960  200) mi3/ 2

1h 3600 s

TC  51.2 h 

PROBLEM 11.161 The oscillation of rod OA about O is defined by the relation   3/ sin  t  where and t are expressed in radians and seconds, respectively. Collar B slides along the rod so that its distance from O is r  6(1  e2t ) where r and t are expressed in inches and seconds, respectively. When t  1 s, determine (a) the velocity of the collar, (b) the acceleration of the collar, (c) the acceleration of the collar relative to the rod.

SOLUTION Calculate the derivatives with respect to time. 3

r  6  6e 2t in.



r  12e2t in/s

  3cos t rad/s

 r  24e 2t in/s 2

  3 sin  t rad/s 2



sin  t rad

At t  1 s,

(a)

3

r  6  6e 2  5.1880 in.



r  12e 2  1.6240 in/s

  3cos   3 rad/s

 r  24e2  3.2480 in/s 2

  3 sin   0



sin   0

Velocity of the collar.

v  rer  re  1.6240 er  (5.1880)(3)e v  (1.624 in/s)er  (15.56 in/s)e 

(b)

Acceleration of the collar.

a  ( r  r2 )er  (r  2r)e

 [3.2480  (5.1880)(3)2 ]er  (5.1880)(0)  (2)(1.6240)(3)]e (49.9 in/s2 )er  (9.74 in/s 2 )e  (c)

Acceleration of the collar relative to the rod. a B /OA   re r

a B /OA  (3.25 in/s2 )er 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.162 The path of a particle P is a limaçon. The motion of the particle is defined by the relations r  b  2  cos  t  and    t, where t and  are expressed in seconds and radians, respectively. Determine (a) the velocity and the acceleration of the particle when t  2 s, (b) the value of  for which the magnitude of the velocity is maximum.

SOLUTION Differentiate the expressions for r and  with respect to time. r  b  2  cos  t  , r    b sin  t ,  r    2b cos  t

   t,    ,   0 sin t  0,

(a) At t  2 s,

r  3b,

r  0,

cos t  1

 r    2b,

  2 rad,

   rad/s

v  r  3 b,

vr  r  0 ,

v  3 be 

ar   r  r 2    2b   3b   2   4 2b

a  r  2r  0,

a   4 2ber 

(b) Values of  for which v is maximum. vr  r    b sin  t v  r   b  2  cos  t   

2 v 2  vr2  v 2   2b 2 sin 2  t   2  cos  t    

  2b 2 sin 2  t  4  4cos  t  cos 2  t    2b 2  5  4cos  t  v

is maximum when ut

cos  t  1

   t,

or hence

 t  0,

2 ,

4 ,

6 , etc

  2N , N  0, 1, 2,  

PROBLEM 11.163 During a parasailing ride, the boat is traveling at a constant 30 km/hr with a 200 m long tow line. At the instant shown, the angle between the line and the water is 30º and is increasing at a constant rate of 2º/s. Determine the velocity and acceleration of the parasailer at this instant.

SOLUTION Given:

v B  30i km/hr  8.333i m/s

eˆ

aB  0

eˆr

r  200 m, r  0,  r 0

  30

r

P

  2 / s  0.0349 rad/s



=0 Relative Motion relations:

B

vP  vB  vP/ B aP  aB  aP / B

Using Radial and Transverse components:

v P / B  rer  re r  r2 er  r  2r e a P / B  



Substitute in known values:







v P / B  6.981e m/s a P / B  0.2437er m/s2

Change to rectangular coordinates:

v P / B  6.981*sin 30i  6.981*cos 30 j m/s =3.491i  6.046 j m/s a P / B  0.2437 *   cos 30  i  0.2437 *sin 30 j m/s 2  0.2111i  0.1219 j m/s 2 Substitute into Relative Motion relations:

v P  8.33i  3.491i  6.046 j m/s =11.824i  6.046 j m/s a P  0  0.2111i  0.1219 j m/s 2  0.2111i  0.1219 j m/s 2

Velocity: Acceleration:

vP  13.280 m/s

27.08 

aP  0.2437 m/s

30.00 

2

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.164 Some parasailing systems use a winch to pull the rider back to the boat. During the interval when  is between 20º and 40º, (where t = 0 at  = 20º) the angle increases at the constant rate of 2 º/s. During this time, / 600 , where r and t are expressed in ft and s, the length of the rope is defined by the relationship respectively. Knowing that the boat is travelling at a constant rate of 15 knots (where 1 knot = 1.15 mi/h), (a) plot the magnitude of the velocity of the parasailer as a function of time (b) determine the magnitude of the acceleration of the parasailer when t = 5 s.

SOLUTION Given:

v B  15i knots*

1.15mph 5280 ft / mi *  25.30i ft/s knot 3600s / hr

aB  0

eˆ

eˆr

1 5 r  600  t 2 m 8 20    40   2 / s  0.0349 rad/s

r P

 B

=0 Take Derivatives of r(t):

Relative Motion relations:

5 32 t m/s 16 15 1  r   t 2 m/s 2 32 r  

vP  vB  vP/ B aP  aB  aP / B

Using Radial and Transverse components:

v P / B  rer  re a P / B   r  r2 er  r  2r e









Change velocity to rectangular coordinates:

v P / B  r( cos  i + sin  j)  r  sin  i  cos  j

Substitute into Relative Motion relations:

      r  r  e   r  2r  e



v P  vB  r cos   r sin  i  r sin   r cos  j aP Note that:

2

v P  vPx i  vPy j a P  ar er  a e

r



PROBLEM 11.164 (Continued)

vPx  vB  r cos   r sin  v  r sin   r cos 

Where:

Py

ar   r  r 2 a  r  2r 

2 2 v P  vPx  vPy

Magnitude of velocity:

a P  ar2  a2 (a) Plot of Velocity as a function of theta: Magnitude of Velocity as Theta Changes 47 46 45

Velocity, (ft/s)

44 43 42 41 40 39 38 37 20

22

24

26

28

30 32 theta, (deg)

34

36

38

40

(b) Magnitude of Acceleration at t=5s:

r  600 

1 52  5  593.012 m 8

3 5  5 2  -3.494 m/s 16 1 15  r    5 2  -1.048 m/s 2 32

r  

    r  2r   0.2439 m/s

a r   r  r 2  1.771 m/s 2 a

2

a P  1.7712  0.24392 aP  1.787 m/s 2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.165 As rod OA rotates, pin P moves along the parabola BCD. Knowing that the equation of this parabola is r  2b /(1  cos  ) and that   kt , determine the velocity and acceleration of P when (a)   0, (b)   90.

SOLUTION 2b   kt 1  cos kt 2bk sin kt   k   0 r  (1  cos kt ) 2 2bk  [(1  cos kt )2 k cos kt  (sin kt )2(1  cos kt )(k sin kt )] r 4 (1  cos kt ) r

(a)

When   kt  0: 2bk 1 [(2) 2 k (1)  0]  bk 2 4 2 (2)

r b

r  0

 r

 0

  k

  0

v  r  bk

vr  r  0

1 1  ar   r  r 2  bk 2  bk 2   bk 2  2 2   a  r  2r  b(0)  2(0)  0  (b)

v  bk e  1 a   bk 2e r  2

When   kt  90°: r  2b

r  2bk

  90

  k

vr  r  2bk

2bk [0  2k ]  4bk 2 19   0  r

v  r  2bk

ar   r  r2  4bk 2  2bk 2  2bk 2 a  r  2r  2b(0)  2(2bk )k  4bk 2 

v  2bk er  2bk e 

a  2bk 2er  4bk 2e 

PROBLEM 11.166 The pin at B is free to slide along the circular slot DE and along the rotating rod OC. Assuming that the rod OC rotates at a constant rate  , (a) show that the acceleration of pin B is of constant magnitude, (b) determine the direction of the acceleration of pin B.

SOLUTION From the sketch:

r  2b cos  r  2b sin   Since   constant,   0

 r  2b cos 2 ar   r  r 2  2b cos   2  (2b cos  ) 2 a  4b cos   2 r

a  r  2r  (2b cos  )(0)  2(2b sin  ) 2 a  4b sin   2 

a  ar2  a2  4b 2 ( cos  ) 2  ( sin  ) 2 a  4b 2

Since both b and  are constant, we find that a  constant 

  tan 1

 4b sin   2 a  tan 1  2 ar  4b cos  

  

  tan 1 (tan  )   Thus, a is directed toward A 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.167 To study the performance of a racecar, a high-speed camera is positioned at Point A. The camera is mounted on a mechanism which permits it to record the motion of the car as the car travels on straightway BC. Determine (a) the speed of the car in terms of b,  , and , (b) the magnitude of the acceleration in terms of b,  , , and .

SOLUTION (a)

We have

r

b cos 

Then

r 

b sin  cos 2 

We have

v 2  vr2  v2  (r) 2  (r)2 2

or

 b sin    b 2       2  cos    cos    b22 b22  sin 2   1    4 cos 2  cos 2  cos  b v cos 2

For the position of the car shown,  is decreasing; thus, the negative root is chosen.

v

b  cos 2

Alternative solution. From the diagram or or

   

r  v sin 

b sin   v sin  cos 2 v

b  cos 2

PROBLEM 11.167 (Continued) (b)

For rectilinear motion

a

dv dt

Using the answer from Part a

v Then

a

b cos 2

d  b   dt  cos 2

 b

  

 cos 2  (2 cos  sin  ) cos 4 a

or

b (  2 2 tan  )  cos 2

Alternative solution From above Then

Now where

and

b b sin  r  cos  cos 2  ( sin    2 cos  )(cos 2 )  ( sin  )(2 cos  sin  )  r b cos 4   sin   2 (1  sin 2  )   b   2 cos3  cos   r

a 2  ar2  a2   sin   2 (1  sin 2  )  b 2  ar   r  r 2  b   2 cos 2  cos   cos  b   2 2 sin 2     sin     cos   cos 2  b sin   ar  (  2 2 tan  ) 2 cos  a  r  2r  

Then

b b 2 sin  2 cos  cos 2

b cos   (  2 tan  ) cos 2

a

b (  2 2 tan  )[(sin  )2  (cos  ) 2 ]1/ 2 2 cos 

For the position of the car shown,  is negative; for a to be positive, the negative root is chosen. b a (  2 2 tan  )  cos 2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.168 After taking off, a helicopter climbs in a straight line at a constant angle  . Its flight is tracked by radar from Point A. Determine the speed of the helicopter in terms of d,  ,  , and .

SOLUTION From the diagram r d  sin (180   ) sin (    ) or

d sin   r (sin  cos   cos  sin  ) tan  tan  cos   sin 

or

rd

Then

r  d tan 

( tan  sin   cos  )   (tan  cos   sin  )2

tan  sin   cos   d tan  (tan  cos   sin  )2 From the diagram

vr  v cos (    )

where

vr  r

Then

tan  sin   cos  d tan   v(cos  cos   sin  sin  ) (tan  cos   sin  ) 2  v cos  (tan  sin   cos  )

v

or Alternative solution. We have

v 2  vr2  v2  (r)2  (r)2

d tan  sec   (tan  cos  sin  )2

PROBLEM 11.168 (Continued)

Using the expressions for r and r from above

 tan  sin   cos   v   d tan   (tan  cos   sin  )2  

2

1/ 2

or

 (tan  sin   cos  )2  d tan   1 v  2 (tan  cos   sin  )  (tan  cos   sin  )  1/ 2

  tan 2   1 d tan    2 (tan  cos   sin  )  (tan  cos   sin  ) 

Note that as  increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen.

v

d tan  sec   (tan  cos   sin  ) 2

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.169 At the bottom of a loop in the vertical plane, an airplane has a horizontal velocity of 315 mi/h and is speeding up at a rate of 10 ft/s2. The radius of curvature of the loop is 1 mi. The plane is being tracked by radar at O. What are the recorded values of r,  r ,  and  for this instant?

SOLUTION Geometry. The polar coordinates are r  (2400) 2  (1800) 2  3000 ft

Velocity Analysis.

 1800    36.87  2400 

  tan 1 

v  315 mi/h  462 ft/s

vr  462cos   369.6 ft/s v  462sin   277.2 ft/s vr  r v  r

 

r  370 ft/s 

v 277.2  3000 r

  0.0924 rad/s  Acceleration analysis.

at  10 ft/s 2 an 

v2





(462) 2  40.425 ft/s 2 5280

PROBLEM 11.169 (Continued)

ar  at cos   an sin   10 cos 36.87  40.425 sin 36.87  32.255 ft/s 2 a  at sin   an cos   10 sin 36.87  40.425 cos 36.87  26.34 ft/s 2 ar   r  r 2  r  ar  r 2

 r  32.255  (3000)(0.0924)2

a  r  2r a 2r     r r 26.34 (2)(369.6)(0.0924)   3000 3000

 r  57.9 ft/s 2 

  0.0315 rad/s 2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.170 Pin C is attached to rod BC and slides freely in the slot of rod OA which rotates at the constant rate  . At the instant when   60, determine (a) r and , (b)  r and . Express your answers in terms of d and  .

SOLUTION

Looking at d and  as polar coordinates with d  0, v  d   d ,

vd  d  0

a  d   2d   0,

(a)

Velocity analysis:

ad  d  d  2  d  2

r  d 3 for angles shown.

Geometry analysis:

Sketch the directions of v, er and e. vr  r  v  er  d  cos120 1 r   d   2

v  r  v  e  d  cos30

  (b)

Acceleration analysis:

d 23 d  cos 30  r d 3

 

1  2

Sketch the directions of a, er and e. ar  a  er  a cos150  

3 d 2 2

3  r  r2   d 2 2  r 

3 3 1  d  2  r 2   d 2  d 3    2 2 2 

2

 r 

3 d 2  4

1 a  a  e  d  2 cos120   d  2 2 a  r  2r 

1   (a  2r)  r

1 3d

 1  1  1   2   2 d   (2)   2 d   2       

  0 

PROBLEM 11.171 For the racecar of Problem 11.167, it was found that it took 0.5 s for the car to travel from the position   60 to the position   35. Knowing that b  25 m, determine the average speed of the car during the 0.5-s interval.

PROBLEM 11.167 To study the performance of a racecar, a high-speed camera is positioned at Point A. The camera is mounted on a mechanism which permits it to record the motion of the car as the car travels on straightway BC. Determine (a) the speed of the car in terms of b,  , and , (b) the magnitude of the acceleration in terms of b,  , , and .

SOLUTION From the diagram: r12  25 tan 60  25 tan 35  25.796 m Now

vave 

r12 t12

25.796 m 0.5 s  51.592 m/s 

or

vave  185.7 km/h 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.172 For the helicopter of Problem 11.168, it was found that when the helicopter was at B, the distance and the angle of elevation of the helicopter were r  3000 ft and   20, respectively. Four seconds later, the radar station sighted the helicopter at r  3320 ft and   23.1. Determine the average speed and the angle of climb  of the helicopter during the 4-s interval.

PROBLEM 11.168 After taking off, a helicopter climbs in a straight line at a constant angle  . Its flight is tracked by radar from Point A. Determine the speed of the helicopter in terms of d,  ,  , and .

SOLUTION We have

r0  3000 ft r4  3320 ft

0  20  4  23.1

From the diagram:

r 2  30002  33202  2(3000)(3320) cos (23.1  20) or

r  362.70 ft

Now

vave 

r t 362.70 ft  4s  90.675 ft/s

vave  61.8 mi/h 

or Also, or or

r cos   r4 cos  4  r0 cos  0 cos  

3320 cos 23.1  3000 cos 20 362.70

  49.7 

PROBLEM 11.173 A particle moves along the spiral shown. Determine the magnitude of the velocity of the particle in terms of b,  , and .

SOLUTION 1 2 

Given:

r  be 2

Take time derivative

r  be 2 

Radial and Transverse Components

1 2 

1 2 

vr  r  be 2  1 2 

v  r  be 2 

Magnitude of Velocity

v 2  vr2  v2 2

 1 2    be 2   2  1  2    





1 2 

v  be 2



2



1

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

1/2

 

PROBLEM 11.174 A particle moves along the spiral shown. Determine the magnitude of the velocity of the particle in terms of b,  , and .

SOLUTION b

Given:

r 

Take time derivative

r  

Radial and Transverse Components

vr  r  

2 2b  

3

v  r 

Magnitude of Velocity

2b  

3

b  

2

v 2  vr2  v2  

4b 2  2 b 2  2   4 6



b

2

6



 4    2

2

v

b



3

4    2

12

 

PROBLEM 11.175 A particle moves along the spiral shown. Knowing that  is constant and denoting this constant by , determine the magnitude of the acceleration of the particle in terms of b,  , and  .

SOLUTION 1 2 

r  be 2   

Given:

Take time derivatives

1 2 

r  be 2  1 2 

 r  be 2

Radial Component of Acceleration

1 2 

 be 2

 

2

  2     2  

 

2

  

     

a  r  2r 

1 2  be 2 



1 2  be 2

  



1 2  2be 2  2

  2 2   

  0

and 1 2 

ar  be 2 Magnitude of Acceleration

  2   

2

ar   r  r 2 1 2 

Substitute Given Values

 

  

 be 2

Transverse Component of Acceleration

  0

and

 2

and

1 2 

a  be 2

 2  2

a 2  ar2  a2 2

 1 2    be 2   4  4 2  4    





a

1 2  be 2 



2

4



Copyright © McGraw-Hill Education. Permission required for reproduction or display.

1 2  2

PROBLEM 11.176 A particle moves along the spiral shown. Knowing that  is constant and denoting this constant by , determine the magnitude of the acceleration of the particle in terms of b,  , and  .

SOLUTION b

Given:

r 

Take time derivatives

r    r 

Radial Component of Acceleration

2b  

3

2b  6b  2   4 3





ar   r  r 2  

Transverse Component of Acceleration

,    and   0

2

2b  6b  2 b   4   2  2 3



b



4





 2  6

2

  2 2



a  r  2r b   2b     2    3   2 2     b  3   4 2 



 &&&a d 



0

Substitute Given Values

ar  Magnitude of Acceleration

b



4

 6    2

2

and

a  

4b

3

2

a 2  ar2  a2  

b2

8 b2

8

36  12 36  4

2

2



  4 4 

16b 2

6

4



  4 4

a

b



4

36  4

2

 4



12

 2

PROBLEM 11.177 The motion of a particle on the surface of a right circular cylinder is defined by the relations R  A,   2 t , and z  B sin 2 nt , where A and B are constants and n is an integer. Determine the magnitudes of the velocity and acceleration of the particle at any time t.

SOLUTION RA R  0

  2 t

z  B sin 2 nt

  2

z  2 n B cos 2 nt

  0 R

  0

 z  4 2 n 2 B sin 2 nt

Velocity (Eq. 11.49)

v  R e R  Re  zk v

 A(2 )e  2 n B cos 2 nt k v  2 A2  n 2 B 2 cos 2 2 nt 

Acceleration (Eq. 11.50)

 - R2 )e  ( R  2 R)e   a  (R zk R  a  4 2 Aek  4 2 n2 B sin 2 nt k a  4 2 A 2  n 4 B 2 sin 2 2 nt 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.178 Show that r  h sin  knowing that at the instant shown, step AB of the step exerciser is rotating counterclockwise at a constant rate .

SOLUTION From the diagram

r 2  d 2  h 2  2dh cos  Then Now or

2rr  2dh sin  r d  sin  sin  r

d sin  sin 

Substituting for r in the expression for r

 d sin      r  dh sin   sin   or

r  h sin 

Q.E.D.

Alternative solution. First note

  180  (   )

Now

v  v r  v  rer  re

With B as the origin

vP  d

(d  constant  d  0)



PROBLEM 11.178 (Continued)

With O as the origin

(vP ) r  r

where

(vP )r  vP sin 

r  d sin 

Then

h d  sin  sin 

Now

d sin   h sin 

or substituting

r  h sin 

Q.E.D.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.179 The three-dimensional motion of a particle is defined by the relations R  A(1  et ),   2 t , and z  B(1  et ). Determine the magnitudes of the velocity and acceleration when (a) t  0, (b) t  .

SOLUTION R  A(1  et ) R  Aet    Aet R

  2 t

z  B(1  et )

  2

z  Bet

  0

 z   Bet

Velocity (Eq. 11.49)

v  R e R  Re  zk v  Aet e R  2 A(1  et )e  Bet k (a)

When

t  0: et  e0  1;

(b)

When

t   : et  e  0

v  A2  B 2 

v  Ae R  Bk v  2 Ae

v  2 A 

Acceleration (Eq. 11.50)

  R1 )e  ( R  2 R)e   a  (R zk R 

 [ Aet  A(1  et )4 2 ]e R  [0  2 Aet (2 )]e  Bet k (a)

When

t  0: et  e0  1 a   Ae R  4 Ae  B k

a  A2  (4 A)2  B 2 (b)

When

a  (1  16 2 ) A2  B 2 

t   : et  e  0

a  4 2 Ae R

a  4 2 A 

PROBLEM 11.180* For the conic helix of Problem 11.95, determine the angle that the osculating plane forms with the y axis.

PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r  (Rt cos nt)i  ctj  (Rt sin nt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.)

SOLUTION First note that the vectors v and a lie in the osculating plane. Now

r  ( Rt cos n t )i  ctj  ( Rt sin n t )k

Then

v

dr  R(cos n t  n t sin n t )i  cj  R(sin n t  n t cos n t )k dt

and

a

dv dt





 R n sin n t  n sin n t  n2t cos n t i





 R n cos n t  n cos n t  n2 t sin n t k  n R[(2sin n t  n t cos n t )i  (2 cos n t  n t sin n t )k ] It then follows that the vector ( v  a) is perpendicular to the osculating plane.

i j k ( v  a)  n R R(cos n t  n t sin n t ) c R(sin n t  n t cos n t ) (2sin n t  n t cos n t ) 0 (2cos n t  n t sin n t )  n R{c(2 cos n t  n t sin n t )i  R[ (sin n t  n t cos n t )(2sin n t  n t cos n t )  (cos n t  n t sin n t )(2 cos n t  n t sin n t )]j  c (2sin n t  n t cos n t )k





 n R c(2 cos n t  n t sin n t )i  R 2  n2t 2 j  c(2sin n t  n t cos n t )k   

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.180* (Continued)

The angle  formed by the vector ( v  a) and the y axis is found from

cos   Where

( v  a)  j | ( v  a) || j |

|j| 1



( v  a)  j  n R 2 2  n2 t 2





|( v  a) |  n R c 2 (2 cos n t  n t sin n t ) 2  R 2 2  n2t 2 



2

1/ 2

 c 2 (2sin n t  n t cos n t ) 2 





1/ 2 2



 

 n R c 2 4  n2 t 2  R 2 2  n2t 2  Then

   R c  4   t   R  2   t     R  2   t   c 4   t  R 2   t        n R 2 2  n2t 2

cos  

2

2 2 n

n

2

2 2 n

2

1/ 2

2 2 n

2

2 2 n

2

2 2 n

1/ 2

2

The angle  that the osculating plane forms with y axis (see the above diagram) is equal to

    90 Then

cos   cos (   90)  sin   sin  

Then

or

tan  

    R  2   t  

 R 2  n2 t 2



c 2 4   2t 2 n 



R 2  n2t 2

2

2 2 n

2

1/ 2



c 4  n2 t 2



 R 2  n2t 2   tan   c 4   2t 2 n  1

    

PROBLEM 11.181* Determine the direction of the binormal of the path described by the particle of Problem 11.96 when (a) t  0, (b) t   /2 s.

SOLUTION





r  ( At cos t )i  A t 2  1 j  ( Bt sin t )k

Given:

r  ft, t  s; A  3, B  1 First note that eb is given by eb 

va |va |





Now

r  (3t cos t )i  3 t 2  1 j  (t sin t )k

Then

v

dr dt 3t

 3(cos t  t sin t )i 

j  (sin t  t cos t )k



t



2 dv t 2 1 a j  3( sin t  sin t  t cos t )i  3 t  12 t dt t 1  (cos t  cos t  t sin t )k 3  3(2sin t  t cos t )i  2 j  (2 cos t  t sin t )k (t  1)3/ 2

and

(a)

t2 1

At t  0:

v  (3 ft/s)i a  (3 ft/s2 ) j  (2 ft/s 2 )k

Then

and Then

v  a  3i  (3j  2k )  3(2 j  3k )

| v  a |  3 (2)2  (3)2  3 13 eb 

3(2 j  3k ) 3 13

cos  x  0 or

 x  90



cos  y  

1 13 2

(2 j  3k )

13

 y  123.7

cos  z 

3 13

 z  33.7

Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 11.181* (Continued)

(b)

At t 

Then

 2

s:

  3   3 ft/s  i   ft/s  j  (1 ft/s)k v      2   2 4     24  ft/s 2  j   ft/s 2  k a  (6 ft/s 2 )i   2 3/ 2   (  4)  2

i 3 va   2

6

j 3 2 (  4)1/2 24 2 (  4)3/ 2

k

1 

 2

   3 2 24 3 2      6 i   2 1/ 2 4 (   4)3/2    2(  4)

  j 

  36 18 k   2  2 3/2 1/ 2  (  4)   (  4)  4.43984i  13.40220 j  12.99459k

and

Then

or

| v  a |  [( 4.43984) 2  (13.40220) 2  (12.99459) 2 ]1/ 2  19.18829

1 (4.43984i  13.40220 j  12.99459k ) 19.1829 4.43984 13.40220 12.99459 cos  x   cos  y   cos  z  19.18829 19.18829 19.18829

eb 

 x  103.4

 y  134.3

 z  47.4



PROBLEM 11.182 The motion of a particle is defined by the relation x  2t 3  15t 2  24t  4, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

SOLUTION x  2t 3  15t 2  24t  4 dx  6t 2  30t  24 dt dv  12t  30 a dt v

so

(a)

0  6t 2  30t  24  6 (t 2  5t  4)

Times when v  0.

(t  4)(t  1)  0 (b)

t  1.00 s, t  4.00 s 

Position and distance traveled when a  0.

a  12t  30  0

t  2.5 s

x2  2(2.5)3  15(2.5)2  24(2.5)  4

so

x  1.50 m 

Final position For

0  t  1 s, v  0.

For

1 s  t  2.5 s, v  0.

At

t  0,

At

t  1 s, x1  (2)(1)3  (15)(1)2  (24)(1)  4  15 m

x0  4 m.

Distance traveled over interval: x1  x0  11 m For

1 s  t  2.5 s,

v0

Distance traveled over interval | x2  x1 |  |1.5  15 |  13.5 m Total distance:

d  11  13.5

d  24.5 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.183 A drag car starts from rest and starts down the racetrack with and acceleration defined by a = 50 – 10 t , where a and t are in m/s2 and s, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a= - 0.02v2, where v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to 10 m/s, (b) the total distance the car travels during this time.

SOLUTION Given:

a1 2  50  10t m/s2 for 0  v  125 m/s a23  0.02v 2 after v  125 m/s

Find v(t) as car speeds up

dv  dv  adt dt

a v t 



t

dv 

0

  50  10t dt 0

v  t   50t  5t 2 Find time to reach v=125 m/s

125  50t  5t 2

Rearrange and solve for t:

t22  10t2  25  0

 t2  5 2  0 t2  5 s

dx  vdt

Find distance to reach-125 m/s: x2

t2

 dx    50t  5t dt 2

0

0

5 x2  25t 2  t 3 3 x2  416.7 m

5

|

0

(a) Find total time to slow down 10 m/s:

a

dv dv  dt  dt a

t3

10

dv v2 125

 dt  50  5

t3  5   

50 10 | v 125

t3  9.6 s 

PROBLEM 11.183 (Continued) (b) Find total distance to slow down to 10 m/s

adx  vdv  dx  x3

vdv a

v3

dv v v2

 dx  50 

x2

10

x3  416.7  50ln v|

125

x3  543.0 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.184 A particle moves in a straight line with the acceleration shown in the figure. Knowing that the particle starts from the origin with v0  2 m/s, (a) construct the v t and x t curves for 0  t  18 s, (b) determine the position and the velocity of the particle and the total distance traveled when t  18 s.

SOLUTION Compute areas under a  t curve. A1  (0.75)(8)  6 m/s A2  (2)(4)  8 m/s A3  (6)(6)  36 m/s v0  2 m/s v8  v0  A1  8 m/s v12  v8  A2  0

v18  v12  A3

v18  36 m/s 

Sketch v t curve using straight line portions over the constant acceleration periods. Compute areas under the v t curve. 1 (2  8)(8)  40 m 2 1 A5  (8)(4)  16 m 2 1 A6  (36)(6)  108 m 2

A4 

x0  0 x8  x0  A4  40 m x12  x8  A5  56 m

x18  x12  A6 Total distance traveled  56  108

x18  52 m  d  164 m 

PROBLEM 11.185 The velocities of commuter trains A and B are as shown. Knowing that the speed of each train is constant and that B reaches the crossing 10 min after A passed through the same crossing, determine (a) the relative velocity of B with respect to A, (b) the distance between the fronts of the engines 3 min after A passed through the crossing.

SOLUTION (a)

v B  v A  v B/A

We have

The graphical representation of this equation is then as shown. Then

vB2 /A  662  482  2(66)(48) cos 155

or

vB/A  111.366 km/h

and

48 111.366  sin  sin 155

  10.50

or

v B/A  111.4 km/h (b)

10.50° 

First note that at t  3 min, A is (66 km/h) at t  3 min, B is (48 km/h) Now

 603   3.3 km west of the crossing.

 607   5.6 km southwest of the crossing. rB  rA  rB/A

Then at t  3 min, we have

rB2/A  3.32  5.62  2(3.3)(5.6) cos 25 or

rB/A  2.96 km 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.186 Knowing that slider block A starts from rest and moves to the left with a constant acceleration of 1 ft/s2, determine (a) the relative acceleration of block A with respect to block B, (b) the velocity of block B after 2 s.

SOLUTION Given:

aA  1 ft/s2

From the diagram can write an expression for the length of the cable

L  x A  3 yB  constants Differentiate:

0  x A  3 y B 0  v A  3vB

Differentiate again: Find the acceleration of B:

Write accelerations as vectors:

0  a A  3aB

1 aB   aA 3 1 2   ft/s 3

a A  1i ft/s2 1 a B  j ft/s2 3

(c) Find relative accelerations:

a A/ B  a A  a B 1  1i  j ft/s2 3 a A/ B  1.054 ft/s2

(d) Find speed at t=2 s:

18.43 

vB   vB o  aB t 0 

1  2 3

2 m/s 3

vb  0.667 m/s  

PROBLEM 11.187 Collar A starts from rest at t  0 and moves downward with a constant acceleration of 7 in./s2 . Collar B moves upward with a constant acceleration, and its initial velocity is 8 in./s. Knowing that collar B moves through 20 in. between t  0 and t  2 s, determine (a) the accelerations of collar B and block C, (b) the time at which the velocity of block C is zero, (c) the distance through which block C will have moved at that time.

SOLUTION 

From the diagram



 y A  ( yC  y A )  2 yC  ( yC  yB )  constant Then

2v A  vB  4vC  0

(1)

and

2a A  aB  4aC  0

(2)

(v A ) 0  0

Given:

(a A )  7 in./s 2 ( v B )0  8 in./s  

a B  constant

y  (yB )0  20 in.

At t  2 s (a)

y B  ( y B ) 0  (vB ) 0 t 

We have At t  2 s:

20 in.  (8 in./s)(2 s) 

aB  4 in./s2

1 aB t 2 2 1 aB (2 s) 2 2

or

a B  2 in./s2 

Then, substituting into Eq. (2)

2(7 in./s 2 )  (2 in./s2 )  4aC  0 aC  3 in./s 2

or

aC  3 in./s 2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.187 (Continued)

(b)

Substituting into Eq. (1) at t  0

2(0)  (8 in./s)  4(vC )0  0 or (vC )0  2 in./s  vC  (vC )0  aC t

Now

(c)

When vC  0:

0  (2 in./s)  (3 in./s 2 )t

or

t

yC  ( yC )0  (vC )0 t 

We have At t 

2 3

2 s 3

s:

t  0.667 s  1 aC t 2 2

2  1 2  yC  ( yC )0  ( 2 in./s)  s   (3 in./s 2 )  s  3  2 3 

 0.667 in.

or

2

y C  (y C )0  0.667 in. 

PROBLEM 11.188 A golfer hits a ball with an initial velocity of magnitude v0 at an angle  with the horizontal. Knowing that the ball must clear the tops of two trees and land as close as possible to the flag, determine v0 and the distance d when the golfer uses (a) a six-iron with   31, (b) a five-iron with   27.

SOLUTION The horizontal and vertical motions are x t cos  1 1 y  (v0 sin  )t  gt 2  x tan   gt 2 2 2 x  (v0 cos  )t

v0 

or

(1)

2( x tan   y) g

or

t2 

At the landing Point C:

yC  0,

And

xC  (v0 cos  )t 

(2) t 

2v0 sin  g

2v02 sin  cos  g

(3)

(a)   31 To clear tree A:

x A  30 m, y A  12 m

From (2),

t A2 

From (1),

(v0 ) A 

To clear tree B:

2(30 tan 31  12)  1.22851 s 2 , 9.81

t A  1.1084 s

30  31.58 m/s 1.1084 cos 31

xB  100 m,

yB  14 m

From (2),

(tB )2 

2(100 tan 31  14)  9.3957 s 2 , 9.81

From (1),

(v0 ) B 

100  38.06 m/s 3.0652 cos 31

The larger value governs,

v0  38.06 m/s

From (3),

xC 

tB  3.0652 s

v0  38.1 m/s 

(2)(38.06) 2 sin 31 cos 31  130.38 m 9.81

d  xC  110

d  20.4 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.188 (Continued)

(b)

  27 By a similar calculation,

t A  0.81846 s, tB  2.7447 s, v0  41.138 m/s xC  139.56 m,

(v0 ) A  41.138 m/s, (v0 ) B  40.890 m/s, v0  41.1 m/s  d  29.6 m 

PROBLEM 11.189 As the truck shown begins to back up with a constant acceleration of 4 ft/s2 , the outer section B of its boom starts to retract with a constant acceleration of 1.6 ft/s2 relative to the truck. Determine (a) the acceleration of section B, (b) the velocity of section B when t  2 s.

SOLUTION For the truck,

a A  4 ft/s 2

For the boom,

a B/ A  1.6 ft/s2

50

(a) a B  a A  a B/ A Sketch the vector addition. By law of cosines:

aB2  a A2  aB2/ A  2a AaB/ A cos 50  42  1.62  2(4)(1.6) cos 50 aB  3.214 ft/s2 Law of sines:

sin  

aB/ A sin 50 aB

  22.4, (b)



1.6 sin 50  0.38131 3.214

a B  3.21 ft/s2

22.4 

v B  (vB )0  aBt  0  (3.214)(2)

v B  6.43 ft/s2

22.4 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.190 A velodrome is a specially designed track used in bicycle racing. If a rider starts from rest and accelerates between points A and B according to the relationship at = (11.46 – 0.01878 v2) m/s2, what is her acceleration at point B?

SOLUTION





at  11.46 – 0.01878v 2 m/s2

Given:

Distance traveled between points A and B:

sB  18.5  r m  18.5  28 *

 2

m

 62.48 m Find velocity at point B:

ads  vdv  ds  62.48

 0

ds 

vB

 11.46 0

vdv a

vdv – 0.01878v 2 vB

1 62.48  ln 11.46  0.01878v 2    2  0.01878 0

2.347  ln 11.46  0.01878vB2   ln 11.46  e0.09211  11.46  0.01878vB2

vB  23.49 m/s Acceleration:

a B  at et 



vB2



 11.46 –

en 0.01878vB2

e

t

2 23.49    e

28

n

a B  1.097et  19.71en m/s 2 

PROBLEM 11.191 Sand is discharged at A from a conveyor belt and falls onto the top of a stockpile at B. Knowing that the conveyor belt forms an angle   25 with the horizontal, determine (a) the speed v0 of the belt, (b) the radius of curvature of the trajectory described by the sand at Point B.

SOLUTION The motion is projectile motion. Place the origin at Point A. Then x0  0 and y0  0. The coordinates of Point B are xB  30 ft and yB  18 ft. Horizontal motion:

Vertical motion:

vx  v0 cos 25

(1)

x  v0 t cos 25

(2)

v y  v0 sin 25  gt y  v0 t sin 25 

(3)

1 2 gt 2

(4)

At Point B, Eq. (2) gives v0 tB 

xB 30   33.101 ft cos 25 cos 25

Substituting into Eq. (4), 1 18  (33.101)(sin 25)  (32.2)t B2 2 t B  1.40958 s (a)

Speed of the belt.

v0 

v0 tB 33.101   23.483 tB 1.40958





v0  23.4 ft/s 

Eqs. (1) and (3) give

vx  23.483cos 25  21.283 ft/s v y  (23.483)sin 25  (32.2)(1.40958)  35.464 ft/s tan 

v y vx

 1.66632

  59.03

v  41.36 ft/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.191 (Continued)

Components of acceleration. a  32.2 ft/s 2

at  32.2sin 

an  32.2cos   32.2cos59.03  16.57 ft/s 2 (b)

Radius of curvature at B.

an 



v2

 v 2 (41.36)2  an 16.57

  103.2 ft 

PROBLEM 11.192 The end Point B of a boom is originally 5 m from fixed Point A when the driver starts to retract the boom with a constant radial acceleration of  r  1.0 m/s 2 and lower it with a constant angular acceleration   0.5 rad/s 2 . At t  2 s, determine (a) the velocity of Point B, (b) the acceleration of Point B, (c) the radius of curvature of the path.

SOLUTION r0  5 m, r0  0,  r  1.0 m/s2

Radial motion.

1 2  rt  5  0  0.5t 2 2 r  r0   rt  0  1.0t r  r0  r0 t 

At t  2 s,

r  5  (0.5)(2) 2  3 m r  ( 1.0)(2)  2 m/s

0  60 

Angular motion.

 3

rad, 0  0,   0.5 rad/s 2

1   0  0.25t 2 2 3   0  t  0  0.5t

  0  0  t 2 





 0  (0.25)(2)2  0.047198 rad  2.70 3   (0.5)(2)  1.0 rad/s

At t  2 s,

Unit vectors er and e .

(a)

Velocity of Point B at t  2 s.

v B  rer  re

 (2 m/s)er  (3 m)(1.0 rad/s)e 

 v B  (2.00 m/s)er  (3.00 m/s)e 



tan  

v 3.0   1.5 vr 2.0

  56.31

v  vr2  v2  (2)2  (3)2  3.6055 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.192 (Continued)

Direction of velocity. et 

v 2er  3e   0.55470er  0.83205e v 3.6055

    2.70  56.31  59.01  (b)

v B  3.61 m/s



59.0° 

Acceleration of Point B at t  2 s.

a B  ( r  r2 )er  (r  2r)e  [1.0  (3)(1)2 ]er  [(3)(0.5)  (2)(1.0)(0.5)]e

a B  (4.00 m/s2 )er  (2.50 m/s 2 )e  tan  

a 2.50   0.625 ar 4.00

  32.00

a  ar2  a2  (4) 2  (2.5) 2  4.7170 m/s 2

    2.70  32.00  29.30 a B  4.72 m/s2 Tangential component:

29.3° 

at  (a  et )et

at  (4er  2.5e )  (0.55470er  0.83205e )et

 [(4)(0.55470)  (2.5)(0.83205)]et

 (0.138675 m/s2 )et  0.1389 m/s2 Normal component:

59.0°

a n  a  at a n  4e r  2.5e  (0.138675)( 0.55470e r  0.83205e )

 (3.9231 m/s 2 )e r  (2.6154 m/s 2 )e

an  (3.9231)2  (2.6154)2  4.7149 m/s2 (c)

Radius of curvature of the path.

an 



v2

 v 2 (3.6055 m/s)2  an 4.7149 m/s

  2.76 m 

PROBLEM 11.193 A telemetry system is used to quantify kinematic values of a ski jumper immediately before she leaves the ramp. According to the system r  500 ft,  r  105 ft/s, r  10 ft/s 2 ,   25,     0.07 rad/s,   0.06 rad/s 2 . Determine (a) the velocity of the skier immediately before she leaves the jump, (b) the acceleration of the skier at this instant, (c) the distance of the jump d neglecting lift and air resistance.

SOLUTION (a)

Velocity of the skier.

(r  500 ft,   25°)

v  vr er  v e  rer  re

 (105 ft/s)er  (500 ft)(0.07 rad/s)e v  (105 ft/s)er  (35 ft/s)e 

Direction of velocity:

v  (105cos 25  35cos 65)i  (35sin 65  105sin 25) j  (109.95 ft/s)i  (12.654 ft/s) j v y 12.654 tan      6.565 vx 109.95 v  (105) 2  (35) 2  110.68 ft/s v  110.7 ft/s

6.57° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.193 (Continued)

(b)

Acceleration of the skier. a  ar e r  a e  ( r  r 2 )er  (r  2r)e ar  10  (500)(0.07) 2  12.45 ft/s 2 a  (500)(0.06)  (2)(105)(0.07)  15.30 ft/s 2

a  (12.45 ft/s 2 )er  (15.30 ft/s 2 )e  a  (12.45)(i cos 25  j sin 25)  (15.30)(i cos 65  j sin 65)  (17.750 ft/s 2 )i  (8.6049 ft/s 2 ) j tan  

ay ax



8.6049 17.750

  25.9

a  (12.45) 2  (15.30)2  19.725 ft/s 2

a  19.73 ft/s 2

(c)

25.9° 

Distance of the jump d. Projectile motion. Place the origin of the xy-coordinate system at the end of the ramp with the x-coordinate horizontal and positive to the left and the y-coordinate vertical and positive downward. Horizontal motion: (Uniform motion)







 Vertical motion:





x0  0 x0  109.95 ft/s





x  x0  x0 t  109.95t 

(Uniformly accelerated motion)

y0  0 y0  12.654 ft/s

 y  32.2 ft/s 

(from Part a)

(from Part a) 

2

y  y0  y0 t 

1 2  yt  12.654t  16.1t 2  2

At the landing point, x  d cos 30

y  10  d sin 30 or

(1) y  10  d sin 30

(2)

PROBLEM 11.193 (Continued)

Multiply Eq. (1) by sin 30° and Eq. (2) by cos 30° and subtract

x sin 30  ( y  10) cos30  0 (109.95t )sin 30  (12.654t  16.1t 2  10) cos30  0 13.943t 2  44.016t  8.6603  0 t  0.1858 s and 3.3427 s

Reject the negative root.

x  (109.95 ft/s)(3.3427 s)  367.53 ft d

x cos 30

d  424 ft 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.CQ1 A bus travels the 100 miles between A and B at 50 mi/h and then another 100 miles between B and C at 70 mi/h. The average speed of the bus for the entire 200-mile trip is: (a) more than 60 mi/h (b) equal to 60 mi/h (c) less than 60 mi/h

SOLUTION The time required for the bus to travel from A to B is 2 h and from B to C is 100/70  1.43 h, so the total time is 3.43 h and the average speed is 200/3.43  58 mph. Answer: (c) 

PROBLEM 11CQ2 Two cars A and B race each other down a straight road. The position of each car as a function of time is shown. Which of the following statements are true (more than one answer can be correct)? (a) At time t2 both cars have traveled the same distance (b) At time t1 both cars have the same speed (c) Both cars have the same speed at some time t < t1 (d) Both cars have the same acceleration at some time t < t1 (e) Both cars have the same acceleration at some time t1 < t < t2

SOLUTION The speed is the slope of the curve, so answer c) is true. The acceleration is the second derivative of the position. Since A’s position increases linearly the second derivative will always be zero. The second derivative of curve B is zero at the pont of inflection which occurs between t1 and t2. Answers: (c) and (e) 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.CQ3 Two model rockets are fired simultaneously from a ledge and follow the trajectories shown. Neglecting air resistance, which of the rockets will hit the ground first? (a) A (b) B (c) They hit at the same time. (d ) The answer depends on h.

SOLUTION The motion in the vertical direction depends on the initial velocity in the y-direction. Since A has a larger initial velocity in this direction it will take longer to hit the ground. Answer: (b) 

PROBLEM 11.CQ4 Ball A is thrown straight up. Which of the following statements about the ball are true at the highest point in its path? (a) The velocity and acceleration are both zero. (b) The velocity is zero, but the acceleration is not zero. (c) The velocity is not zero, but the acceleration is zero. (d) Neither the velocity nor the acceleration are zero.

SOLUTION At the highest point the velocity is zero. The acceleration is never zero. Answer: (b) 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.CQ5 Ball A is thrown straight up with an initial speed v0 and reaches a maximum elevation h before falling back down. When A reaches its maximum elevation, a second ball is thrown straight upward with the same initial speed v0. At what height, y, will the balls cross paths? (a)

yh

(b)

y  h/2

(c)

y  h/2

(d)

y  h/2

(e)

y0

SOLUTION When the ball is thrown up in the air it will be constantly slowing down until it reaches its apex, at which point it will have a speed of zero. So, the time it will take to travel the last half of the distance to the apex will be longer than the time it takes for the first half. This same argument can be made for the ball falling from the maximum elevation. It will be speeding up, so the first half of the distance will take longer than the second half. Therefore, the balls should cross above the half-way point. Answer: (b) 

PROBLEM 11.CQ6 Two cars are approaching an intersection at constant speeds as shown. What velocity will car B appear to have to an observer in car A? (a)

(b)

(c)

(d )

(e)

SOLUTION Since vB  vA+ vB/A we can draw the vector triangle and see

v B  v A  v B/A

Answer: (e) 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.CQ7 Blocks A and B are released from rest in the positions shown. Neglecting friction between all surfaces, which figure below best indicates the direction  of the acceleration of block B?

SOLUTION    Since aB  a A  aB/A we get

Answer: (d ) 

PROBLEM 11.CQ8 The Ferris wheel is rotating with a constant angular velocity . What is the direction of the acceleration of Point A? (a) (b) (c) (d ) (e) The acceleration is zero.

SOLUTION The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration pointed upwards. Answer: (b) 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.CQ9 A racecar travels around the track shown at a constant speed. At which point will the racecar have the largest acceleration? (a) A (b) B (c) C (d ) The acceleration will be zero at all the points.

SOLUTION The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration. The normal acceleration will be maximum where the radius of curvature is a minimum that is at Point A. Answer: (a) 

PROBLEM 11.CQ10 A child walks across merry-go-round A with a constant speed u relative to A. The merry-go-round undergoes fixed axis rotation about its center with a constant angular velocity  counterclockwise.When the child is at the center of A, as shown, what is the direction of his acceleration when viewed from above. (a) (b) (c) (d ) (e) The acceleration is zero.

SOLUTION Polar coordinates are most natural for this problem, that is,  a  ( r  r2 )er  (r  2r)e

(1)

From the information given, we know  r  0,   0, r  0,   , r  -u. When we substitute these values into (1), we will only have a term in the  direction. Answer: (d ) 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.