Free Study Material Kinematics

KINEMATICS

POSITION VECTOR If the coordinates of a particle are given by (x2, y2, z2) its position vector with respect to (x1, y1, z1) is given by

r = (x2 - x1) i + (y2 - y1) j + (z2 - z1) k . Usually, position vector with respect to the origin

(0, 0, 0) is specified and is given by r =x i +y j +z k .

DISPLACEMENT Displacement is a vector quantity. It is the shortest distance between the final and initial positions of a particle. If

r

r 1 is the initial position r1

vector and r 2 is the final position vector, the displacement vector is given by

r

r2

r2

r 1.

The magnitude of the displacement is given by

x2

x1 2

y1 2

y2

z2

z1 2

This is nothing but the straight line distance between two points (x 1, y1, z1) and(x2, y2, z2).The displacement is independent of the path taken by the particle in moving from (x1, y1, z1) to (x2, y2, z2) DISTANCE If a particle moves along a curve, the actual length of the path Q( x 2 , y2 , z 2 )

is the distance. Distance is always more than or equal to P(x 1 , y 1 , z 1 ) displacement.

Illustration 1: A car travels along a circular path of radius (50 / ) m with a speed of 10 m/s. Find its displacement and distance after 17.5 sec. Solution: Distance = (speed) time = 10 (17.5) = 175 m Perimeter of the circular path = 2

A

(50/ ) = 100 m

The car covers 1 3 rounds of the path 4

B

If the car starts from A, it reaches B and the displacement is the shortest distance between A and B Displacement =

R2

R2

50 2

2 R

=

m.

INSTANTANEOUS AND AVERAGE VELOCITY If

r is the displacement of the particle in time t, the average velocity is given by V

average

=

r2 t2

r1 t1

r t

= Find value - Initial value. The above definition is valid for any magnitude of But when

large or small.

is infinitesimally small, the instantaneous velocity is obtained.

V

=

instantaneous

Lt t

r dr = t dt

0

In normal notation, velocity refers to the instantaneous velocity.

SPEED Speed =

Dis tan ce time

When the time under consideration is very small, distance becomes equal to the displacement and speed becomes the magnitude of instantaneous velocity. Speed is represented only by its magnitude where as velocity is represented by magnitude as well as direction.

INSTANTANEOUS AND AVERAGE ACCELERATION If

V is the change in velocity in time t, average acceleration is given by

V . t

a average =

When

Lt

becomes infinitesimally small, t

acceleration.

0

V dV = which gives the instantaneous t dt

In normal notation, acceleration refers to the instantaneous acceleration.

dV dt

a

d dt

dr dt

d2 r

It may be noted here that magnitude of

dt 2

=

d2 r dt 2

is not equal to

d2r dt 2

always (as in the case of circular

motion)

Illustration 2: A bus shuttles between two places connected by a straight road with uniform speed of 36 kmph. If it stops at each place for 15 minutes and the distance between the two places is 60 km, find the average values of

(a) Speed

(b) Velocity

(c) acceleration between t = 0 and t = 2 hours and the instantaneous values of (d) Velocity

(e) acceleration at t = 2 hrs.

Solution: Time taken for forward trip =

60 5 = hrs. 36 3

Time of stoppage = 15 min = 0.25 hrs. Time available for return trip = 2 - 5/3 - 0.25 = 1/12 hrs. Distance travelled in the return trip = (36) 1/12 = 3 km.

d

60km d1

(a) Average speed =

Total distance 60 3 = = 31.5 kmph. Total time 2

(b) Average velocity =

Displacement 60 3 = = 28.5 kmph Time 2

(c) Average acceleration =

=

3km

V2 V1 = t

change in velocity Time

36

36 2

= - 36 km/H2 = -

1 m/s2 360

(d) Velocity at t = 2 hours = - 36 kmph (e) Acceleration at t = 2hours = 0 as there is no change in velocity

Illustration 3: A car travels towards North for 10 minutes with a velocity of 60 Kmph, turns towards East and travels for 15 minutes with a velocity of 80 kmph and then turns towards North East and travels for 5 minutes with a velocity of 60 kmph. For the total trip, find (a) distance travelled (b) displacement (c) average speed (d) average velocity and (e) average acceleration.

Solution:

S3

Total time taken = (10 + 15 + 5) min = 1/2 hour

S2

(a) Distance travelled = d1 + d2 + d3 = 60

S1

10 15 5 + 80 + 60 60 60 60

S

= 10 + 20 + 5 = 35 km (b) Displacement S

S1 S 2 S 3

= 10 j + 20 i +5 cos 450 i + 5 sin 45 j = 23.5 i + 13.5 j Magnitude of displacement = (c) Average speed =

13.5 2 ~ 27 km

Total distance travelled Time

(d) Average velocity =

at an angle

23.5 2

35 km = 70 kmph. 1 hr 2

23.5i 13.5 j Displacement = 1 Time 2

with the East given by Tan

Magnitude of average velocity = (e) Average acceleration =

47 2

=

47 i

27 j kmph.

13.5 23.5

27 2 = 54 kmph

change in velocity Time =

Final velocity Initial velocity Time

=

60 cos 45 0 i 60 sin 45 j 2

= 21i at an angle

60 j

9 j km/H2

9 21

with the East given by Tan =

212

Magnitude of average acceleration =

9 2 ~ 23 km/H2 ~ 1.8 x 10-3 m/s2

Illustration 4: A car moving along a circular path of radius R with uniform speed covers an angle

during a given

time. Find its average velocity and average acceleration during this time. Solution: Let V be the speed of the car V=

Distance R = time t

where

Displacement= R 2

R2

is in radians.

2R 2 cos

from the triangle OAB

= 2R sin /2

Diplacemnt Average velocity = = Time

Average acceleration =

V=

V2

2 R sin R V

V2

is small sin ~

2 =

2

2 V 2 sin

2

R

and

2 V sin Average velocity =

R V

V

2

V t

2V 2 cos = 2 V sin

Average acceleration =

2 V sin

V

Change in velocity time

2 V sin

When

2 =

2 =

2V

2

V

Average velocity = Instantaneous velocity for small angular displacements

V

2 V 2 sin Average acceleration =

2

2V 2

R

2

R

V2 = R

Average acceleration = Instantaneous acceleration for small angular displacements.

KINEMATICAL EQUATIONS : ( CONSTANT ACCELERATION )

v = u + at v2 = u2 + 2as s= ut + 1/2 at2

The above equations are valid only for constant acceleration and in a particular direction. u,v and s must be taken with proper sign. Usually the direction of u is taken as positive and the sign of other variables are decided with respect to this direction. Displacement during the nth second is Sn - Sn-1 = u +

a (2n - 1) 2

It may be noted here that this is not the distance travelled in the nth second.

Illustration 5: A particle is vertically projected upwards with an initial velocity of 22.5 m/s. Taking g=10 m/s 2 find (a) velocity (b) displacement (c) distance travelled in t = 4 sec and (d) displacement and distance travelled in 3rd second. Solution: Taking the upward direction positive (a)

v = u + (-g) t = 22.5 - 10 (4) = -17.5 m/s 2

17.5 m/s down wards 2

(b)

s = ut + 1/2 (-g) t

= 22.5 (4) - 1/2 (10) 4 = 10 m

(c)

Time to reach the top most point = t0 and at the top most

d2

point velocity becomes zero. V = u - gt0

0 = 22.5 - 10 (t0)

t0 = 2.25 sec

Distance travelled in 4 sec = d1 + d2 d1 = u t0 - 1/2 g t 02 = 22.5 (2.25) - 1/2 (10) (2.25)2 = 25.3 m d1 can be found from v2 - u2 = 2a S also.

d1

0 - (22.5)2 = 2(-10) d1

d1 =

22.5 2 = 25.3 m 20

d2 can be found from s= ut + 1/2 at2 applied along the down ward direction starting from the top most point d2 = 0 (t - t0) + 1/2 g (t - t0)2

= 1/2 (10) (4 - 2.25)2 = 15.3 m

Distance travelled in 4 sec = 25.3 + 15.3 = 40.6 m Displacement in 4 sec = d1 - d2 = 25.3 - 15.3 = 10 m Displacement can also be found directly by applying S = ut + 1/2 at 2 along the vertical Displacement in 4 sec = 22.5 (4) - 1/2 (10) (4)2 = 10m (d)

3rd second is from t = 2 sec to t = 3 sec. Displacement in the 3rd second = u +

a (2n - 1) 2

= 22.5 -

10 (6 - 1) = -2.5 m 2

When there is no change in the direction of the motion along a straight line, distance will be equal to displacement. When the particle reverses its direction during the time under consideration, distance will be more than the displacement and the time at which the reversal is taking place must be found. When the particle reverses its direction, its velocity becomes zero.

t

d1

Using v = u + at, 0 = 22.5 - 10 (t0)

t0 = 2.25 sec

d = d1 + d2 Using the formula S = ut + 1/2 at2 d1 = [22.5 (2.25) - 1/2 (10) (2.25)2 ] - [22.5 (2) - 1/2 (10) (2)2] = 0.31 m Along the downwards vertical starting from the top d2 = 0 (3 - 2.25) + 1/2 (10) (3 - 2.25)2 = 2.81 m d = 0.31 + 2.81 = 3 .12 m

KINEMATICAL EQUATIONS ( VARIABLE ACCELERATION ) :

When the acceleration is variable, the kinematical equations take the form v=

dx dt

2.25 sec

t

2 sec

d2 t

3 sec

d 2x dt 2

dv a= dt dv dx

a=

dx dt

vdv dx

t

a dt

v=u+ 0

t

t

a dt dt and

x = ut + 0

0

x 2

2

a dx

v -u =2 0

Illustration 6: The position coordinate of a particle moving along a straight line is given by x = 4 t 3-3t2+4t+5. Find (a) Velocity and acceleration as a function of time

(b) Displacement as a function of time (c) the

time at which velocity becomes zero and the acceleration at this time (d) the time at which acceleration becomes zero and the velocity at this time. Solution: (a)

v=

d x x0 ds = dt dt

=

dx Where x0 is the initial position coordinate which is a dt

constant =

a= (b)

d (4t3 - 3t2 + 4t + 5) = 12 t2 - 6t + 4 dt dv d = (12t 2 - 6t + 4) = 24t - 6 dt dt

Displacement = (position coordinate at time t) - (position coordinate at t = 0) = (4t3 - 3t2 + 4t + 5) - (5) = 4t3 - 3t2 + 4t

(c)

When v = 0, 12 t2 - 6t + 4 = 0

t=

3

9 48 12

since this value is imaginary, the velocity never becomes zero.

(d)

When a = 0, 24t – 6 =0

1 this time, V= 12 4

and

t=

1 4

4

2

6

6 1 = units and the velocity of the particle at 24 4

13 units 4

Illustration 7: The velocity of a particle moving in the positive direction of the x axis varies as V =

x where

is a positive constant. Assuming that at the moment t = 0 the particle was located at the point x = 0, find (a) the time dependence of the velocity and the acceleration of the particle

(b) the mean

velocity of the particle averaged over the time that the particle takes to cover the first S meters of the path. Solution: (a)

x

dx v= = dt

x

t

dx

0

dt

x

0

2 2

2

x =

t 2

V=

(b)

and x =

t

and

2

a=

t 4

2 dV = 2 dt

Displacement time

Mean velocity =

Displacement = S, and the time taken for this displacement t =

S

Mean velocity =

2 S

=

S 2

Mean velocity can also be found from the following formulae

v dx v mean =

when v is a function of x

dx

v dt and v mean =

when v is a function of time.

dt

2 S

KINEMATICAL EQUATIONS IN VECTOR FORM ( CONSTANT ACCELERATION )

(i) v

u a t

(ii) v . v u . u

(iii) s

2a. s

ut

1 a t2 2

The above equations are useful in 2 and 3 dimensional motion.

Illustration 8: A particle moving on a horizontal plane has velocity and acceleration as shown in the diagram at time t = 0. Find the velocity and displacement at time’t’. Solution:

y

METHOD - I

a

a = - a cos 45 i - a cos45 j =

V = u

a t =

3 u 2

a 2

2

t

ut

1 a t2 = 2

3 ut 2

a

i -

i +

1 2

The magnitude of the displacement =

a

u 2

2

2 2

2

Sx 2

i

x

a 45 0

a

at

t2

30 0

j

2

3 u 2

The magnitude of the velocity =

S

u

3 u u i + j 2 2

u = u cos300 i + u cos 60 j =

t

j

u 2 u t 2

2

a 2 1 2

t a 2

t2

j = Sx i + Sy j

S y2

u

METHOD - II

30 0

45 0

This can be solved by vector addition method also. It may be noted here that u t

1 2 will be along the direction of u , a t and a t will be along the direction of a 2

V

u

at

Since the angle between u and a t is 1650, the magnitude of the velocity is

a t V

at 2

u2

2 u at cos1650

ut S

1 a t2 2

ut

u t and

Since the angle between

ut

displacement is

2

30 0

1 2 at 2

1 a t 2 is 1650, the magnitude of the 2

2

2 ut

S

1 2 at cos 1650 2

KINEMATICAL EQUATIONS IN RELATIVE FORM ( CONSTANT ACCELERATION ) : When two particles A and B move simultaneously with initial velocities u

VA - VB ;

VAB

SAB

SA SB ; a AB

A

and u B , at any time 't'

aA aB

U AB = U A U B VAB = u AB a AB t S AB

u AB t

1 a AB t 2 2

where X AB means parameter X of A with respect to B. Similarly if r is the position coordinate at time’t’ and r0 is the initial position coordinate at time’t’ = 0,

rA

r0 A

uAt

1 aA t2 2

rB

r0 B

uBt

1 aB t2 2

rAB

r0 AB + u AB t

1 a AB t 2 gives the position coordinate of A with respect to B at 2

any time. rAB gives the distance between A and B at any time’t’. Illustration 9: A loose bolt falls from the roof of a lift of height 'h' moving vertically upward with acceleration 'a'. Find the time taken by the bolt to reach the floor of the lift and the velocity of impact.

450

1 a t2 2

Solution:

S b

hj as the bolt travels a distance 'h' downwards before hitting the floor

a b

ab

a  = (-g j ) - (a j ) = - (g + a) j

u b

ub

u  = u j - u j = 0 as they have the same initial velocity upwards

S b

u b t

-hj =0-

1 a b t 2 2

1 (a + g) t2 j 2

t=

2h a g

Velocity of impact is nothing but the relative velocity of the bolt with respect to the lift vimpact = V b = - (a+g) j

u b

2h a g

a b t =

2h a g

j

Illustration 10: uB

Two particles A and B move on a horizontal figure. If the initial distance of separation between A

60 0

10 m

surface with constant velocities as shown in the

45 0

10 m / s

B

them is 10 m at t=0, find the distance between uA

them at t = 2 sec

10 2 m / s

Solution: Distance between them = r AB Taking the origin at the initial position of A

u AB

r AB

r0 AB

r0 AB

10 i

UA UB

u AB t

1 a AB t 2 2

10 2 cos 45 0 i 10 2 cos 45 0 j

= 5 i - 18.7 j

10 cos 60 0 i

10 cos 30 0 j

a AB r AB

aA

0

aB

10 i

At t = 2 sec ,

5i

18.7 j t 37.4 j and r AB = 37.4 units.

r AB

DISPLACEMENT - TIME GRAPHS

The displacement is plotted along 'y' axis and the time along 'x' axis. The slope of the curve the instantaneous velocity at that point.

The average slope between two points

ds gives dt

s gives the t

average velocity between these points. Rate of change of slope gives the acceleration. If the slope is positive and decreases with time, the particle is under retardation. If the slope is positive and increases with time, the particle is under acceleration, constant slope implies zero acceleration.

Illustration 11: The displacement - time graph of a particle moving along a straight line is given below. Find a) the time at which the velocity is zero

2m x

Semi circle 0

b) the velocity at time t = 1 sec

2

4 t

c) the average velocity between t = 2 sec and t = 4 sec

Solution: (a)

Velocity is zero when the slope is zero which happens at t = 2 sec

(b)

Since any point (x, t) lies on the circle of radius 2 m and centre (2, 0). (x-0)2 + (t - 2)2 = 22

x=

velocity is given by the slope

V=

d dt

4

4

t 22

dx =V dt 1

t 22 2 4

t 22

2t 2

=+

1 3

Since the slope is +ve between t = 0 and t = 2, v =

(c)

Average velocity =

1

m/s

3

Displacement O 2 = = -1 m/s time 4 2

VELOCITY - TIME GRAPH If velocity is plotted on 'y' axis and time is plotted on x axis, the slope of the curve at any point

dv given instantaneous V dt

A3

A1

v gives t

acceleration. The average slope between two points

t

A2

average acceleration. The total area between the curve and the time axis gives distance where as algebraic sum of the areas gives displacement. Distance = A1 + A2 + A3 Displacement = A1 - A2 + A3 The nature of acceleration can be found from the rate of change of slope.

Illustration 12: The velocity time graph of a particle moving along a straight line has the form of a parabola v = (t2 - 6t + 8) m/s . Find (a) the distance travelled between t = 0 second t = 3 sec(b) the velocity of the particle when the acceleration is zero (c) the acceleration of the particle when the velocity is zero (d) the velocity of the particle when the acceleration is zero

Solution: (a)

Distance = area OAB + area BCF which can be obtained by the method of integration. Since at the points B and D, velocity becomes

A

E

v

zero

t2 – 6t + 8 = 0 t = 2 sec and 4sec

2

4 2

A3

A2

C

Since F is in between B and D, the time corresponding to F is

D A4

F

A1 B

O

= 3 sec. Similarly A corresponds to t = 0 and E corresponds to t = 6 sec 2

2

v dt

Area OAB = A1 = 0

t 0

2

6t 8 dt

t3 3

6t 2 2

2

8t

= 0

20 m 3

t

3

v dt

Area BCF = A2 = 2

Distance =

20 3

2 3

t3 3

6t 2 2

8t

2 m 3

22 m 3

(b)

Displacement between t = 3 sec and t = 6 sec = A4 - A3 = A1 - A2 =

(c)

a=

dv = 2t - 6 dt

When v = 0;

t = 2sec and 4 sec

a = 2(2) - 6 and 2(4) - 6 - 2 m/sec2

= (d)

When a = 0,

and 2 m/sec2

2t - 6 = 0 and t = 3 sec

2

v = 3 - 6(3) + 8 = -1 m/sec

PROJECTILE MOTION

T u

At the top most point vy = 0 and

H

vx = u cos From vy = uy + ay t, 0 = usin - gt t=

Time of flight = 2t =

From v y2

R

u sin g

2 u sin g

u y2 = 2ay sy, 0 - (u sin )2 = 2 (-g) H and H =

Range = (Time of flight) (horizontal velocity) =

Range is maximum when

R = H

u 2 sin 2 2g

2 u sin g

= 450 and Rmax =

2 u 2 sin cos g = 4 cot

u 2 sin 2 2g

u2 g

u cos

u 2 sin 2 g

20 3

2 = 6m 3

The velocity of the particle at any time’t’ is given by v

vx i

vy j

v = (ucos ) i + (usin - gt) j

The magnitude of the velocity = If

u cos 2

gt 2

u sin

is the angle made by the velocity at any time’t’ with the horizontal, Tan

u sin gt u cos

=

Taking the origin at the point of projection, the 'x' and 'y' coordinates at any time’t’ is given by x = u cos t

and y = usin t

1 2 gt 2

Eliminating’t’ from x and y

1 x g 2 u cos

x u cos

y = u sin

= x tan -

gx 2 2 u 2 cos2

2

which is the equation of a parabola.

It may be noted here that the velocity of the projectile will be always tangential to its path.

The

equations of projectile motion derived above are valid only for constant acceleration due to gravity 'g'.

Illustration 13: A particle is projected from the horizontal at an inclination of 600 with an initial velocity 20 m/sec. Assuming g = 10 m/sec2 find

(a) the time at which the energy becomes three fourths kinetic and one

fourth potential (b) the angle made by the velocity at that time with the horizontal (c) the x and y coordinates of the particle taking the origin at the point of projection.

Solution: (a)

Let v be the velocity when the given condition is fulfilled

v=

v

3u = 10 2

u cos

i

1 3 1 mv2 = ( mu2) 2 4 2

3 m/sec

u sin

gt j

= 20 cos 600 i + 20 sin 600

10t j = 10 i + 10 3 10 t j

3

v = 10

102 + 10 3

3

Solving t =

3

t= (b)

Tan

2

= 10 3

2

2 sec

2 while rising up and t =

3

2 while coming down

u sin gt u cos

=

= (c)

10 t

10 3

x = ucos

10 3 10

2

2

=+

t

3

= 10

2 = 10

and y = u sin t -

3

= 10

3

3

2 m or 10

3

2 m

1 2 gt 2 2 -5

3

2

2

5m

PROJECTILE MOTION ON AN INCLINED PLANE

Let

be the inclination of the plane and y '

the particle is projected at an angle inclined plane.

y

with the

It is convenient to take the

x'

u

reference frame with x' along the plane and y' perpendicular to the plane.

gcos

will be the

component of the acceleration along the downward perpendicular to the plane and g sin

Along the plane, the kinematical equations take the form

v x'

u x'

a x 't

v x'

u cos

g sin

s x'

ux 't

1 a x't 2 2

s x ' = ucos t 2

g cos x

will be the component of the acceleration along the downward

direction of the inclined plane.

vx' - u x ' 2

g sin

2a x 'Sx '

t

1 gsin t2 2 2 vx' - (ucos )2 = 2 (-g sin ) s x '

Similarly perpendicular to the plane, the kinematical equations take the form

v y'

g cos t

u y'

s y'

u y 't

1 a y' t 2 2

s y'

u sin t

u y2'

2 a y' s y'

v y2'

u sin

v y'

2

a y 't

u sin

v y'

1 g cos 2 2

2

t2

g cos

s y'

Here it may be noted that, When the particle strikes the inclined plane s y ' = 0 When the particle strikes the inclined plane perpendicular to it, s y ' When particle strikes the inclined plane horizontally s y '

0 and v x '

0

0 and v y = 0

Illustration 14: From the foot of an inclined plane of inclination , a y ' projectile is shot at an angle with the inclined plane. Find the relation between

and

u

x'

if the projectile

strikes the inclined plane (a) perpendicular to the plane (b) horizontally

Solution: (a)

Since the particle strikes the plane perpendicularly s y '

1 g cos 2

u sin t -

t=

2 u sin g cos

and

2 u sin g cos

(b)

t2 = 0

u cos g sin

and

t=

u cos

0 and v x '

- g sin t = 0

u cos g sin

2 Tan

= cot

Since the particle strikes the plane horizontally s y ' = 0 and v y u sin t -

t=

1 g cos 2

2 u sin g cos

=

t2 = 0

u sin g

and

u sin ( + ) - gt = 0

2 u sin g cos

u sin g

0

0

2 sin cos

= sin ( + )

CIRCULAR MOTION

When a particle moves in a circle of radius R with constant speed v, its called uniform circular

v

motion.

v

v

V

When the particle covers , the direction of velocity also changes by Change in velocity

v

without change in magnitude.

v will be towards the centre of curvature of the circular path which causes

centripetal acceleration.

is called the angular position (or) angular displacement.

Centripetal acceleration, a r =

v t

The rate of change of angular position is known as angular velocity ( ) Time period of circular motion T =

2 R v

In the same time the particle covers an angle 2 =

2 2 v = T 2 R t=

When

is small sin

=

R v

~

Centripetal acceleration =

from which angular velocity can be found as

v R and

v=

v2

v2

2v 2 cos

= 2v sin

v=v

v = t

v R v

=

v2 R

When speed of the particle continuously changes with time, the tangential acceleration is given by

at =

dv dt

2

The rate of change of angular velocity is called the angular acceleration ( ) since a r and a t are perpendicular to each other, the resultant acceleration is given by a = Angle made by the resultant with radius vector tan =

ar2

at2

at ar

Illustration 15: The speed of a particle in circular motion of radius R is given by v = Rt2. Find the time at which the radial and the tangential accelerations are equal and the distance traveled by the particle up to that moment.

Solution:

ar

at

v2 R

dv = 2Rt dt

t=

1 23

v dt =

Distance travelled = 0

1 23

Rt 3 3

1 23

=

2R 3

RADIUS OF CURVATURE When a particle is moving in a plane a r

v2 where v is the instantaneous velocity and R is R

the radius of curvature at that point. Radius of curvature =

v2 ar

If the path of the particle is given by y = f(x), radius of curvature can also be found from the formula

1 R=

Illustration 16:

dy dx d2y dx 2

2

3 2

A particle is projected with initial velocity ‘u’ at angle

with the horizontal. Find the radius of

curvature at (a) point of projection (b) the top most point.

Solution:

at the point of projection P, v = u and a r = g cos

(a)

R=

v2 u2 = ar g cos

p

at the topmost point T, v = u cos and a r

(b)

T

u

g

v2 u 2 cos2 R= = ar g

SHORTEST DISTANCE OF APPROACH When two particles A and B are moving simultaneously, their position coordinates at any time’t’ are given by (when the accelerations are uniform)

rA

r0 A

1 a A t 2 and r B 2

uA t

1 a B t2 2

r0 B u B t

The distance between them at any time’t’, S = r AB

where r AB

r0 AB u AB t

1 a AB t 2 2

The distance between them becomes minimum when

ds =0 from which the time at which it becomes dt

minimum can be found. Substituting the value of time so obtained in r AB , s min can be found.

Illustration 17: Two ships A and B move with constant velocities as North

shown in the figure. approach between them

Find the closest distance of A

30 o

VA

20 kmph

10 km

O

45 0

20 km

B

VB

East

10

2 kmph

Solution:

r0 A

10 j

VA

20 cos 60 0 i

10i

aA

20 i

r0 B

20 cos 30 j

10 3 j

0

aB

r0 A V A t

rB

= 10 t i + 10 10 3 t j

r AB

rA

r0 B V B t

= (20 + 10 t) i + 10 t j

r B = - 20 i + 10 10 3 t

S = r AB =

20

2

1 2

10 - 10

20

10 10 3 t

3 t - 10 t = 0

10 t

t=

10 t j

10 10 3 t

10 t

2 10 10 3 t

2

1 1

2

dS =0 dt

When the distance between A and B is minimum

2

10 2 cos 45 j

= 10 i + 10 j

0

rA

10 2 cos 45 i

VB

3

10 t

10 3

10 = 0

hr

Substituting this value of time in the expression for S, S min = 20 km

CYCLIC MOVEMENT OF PARTICLES When three or more particles located at the vertices of a polygon of side l move with constant speed V such that particle 1 moves always towards particle 2 and particle 2 moves always towards 3 particle etc., they meet at the centre of the polygon following identical curved paths. Time of meeting =

Initial seperation Velocityof approach

Velocity of approach is the component of the relative velocity along the line joining the particles.

Illustration 18: Six particles located at the six vertices of a hexagon of side l move with constant speeds V such that each particle always targets the particle in front if it. Find the time of meeting and the distance travelled by each particle before they meet

Solution: t=

Initial seperation Velocity of approach

60 0 V

V

 2 = = 0 V V V cos60

V

V

Since they move with constant speed V, the distance travelled by each particle in time t =

2 is V

d = Vt = V

2 =2l V

V

RIVER PROBLEMS

C

B

If V r is the velocity of the river and V b is the velocity of the boat with respect to still water, the resultant

Vr

velocity of the boat V R = V b V r Only the perpendicular component of the resultant velocity helps in crossing the river. Time of crossing, t =

w Vb cos

Vb

A

where 'w' is the width of the river.

The boat crosses the river in the least time when

=0

The parallel component of the resultant velocity determines the drift. Drift is the displacement of the boat parallel to the river by the time the boat crosses the river Drift, x = Vr

Vb sin

w Vb cos

Zero drift is possible only when Vr = Vb sin . When Vr > Vb zero drift is not possible.

Illustration 19:

A river of width 100 m is flowing towards East with a velocity of 5 m/s. A boat which can move with a speed of 20 m/s with respect to still water starts from a point on the South bank to reach a directly opposite point on the North bank. If a wind is blowing towards North East with a velocity of 5

2

m/s, find the time of crossing and the angle at which the boat must be rowed. Solution:

20 sin i

Vb

20 cos j Vw

Vr = 5 i Vw=5

Vb

Vr

45

2 cos 45 i + 5 2 cos 45 j = 5 i + 5 j

V R = Resultant velocity of the boat = V b + V r + V w = ( - 20 sin

+ 5 + 5) i + (20 cos

+ 5) j

For reaching directly opposite point, the component of the resultant velocity parallel to the river must be zero - 20 sin + 10 = 0

sin =

1 and = 300 2

Since time of crossing depends only on the perpendicular component of the resultant velocity.

t=

w 20 cos

5

=

100 20 cos 30 0

5

= 4.48 sec

WORKED OUT OBJECTIVE PROBLEMS

EXAMPLE: 01 A point moves along 'x' axis. Its position at time 't' is given by x 2 = t 2 + 1. Its acceleration at time’t’ is (A)

1 x3

(B)

1 x

1 x2

(C)

t

(D)

x2

Solution: x=

t2

1 ;

dx 1 = (2t) = dt 2 t2 1

t t2

1

t2 x3

t2

2

d x = xt 2

a=

t

1

2 t t2

2

1

1

2

(2 t )

1

=

t2

1

3

=

1 x3

EXAMPLE: 02 A body thrown vertically up from the ground passes the height 10.2m twice at an interval of 10 sec. Its initial velocity was (g = 10 m/s2) (A) 52 m/s

(B) 26 m/s

(C) 35 m/s

(D) 60 m/s

Solution: Displacement is same in both cases s = ut + 1/2 at2 10.2 = ut -

t1 =

u2

u

u

u 2 204 10

u 2 204 u and t2 = 10

u 2 204 10

1 (10) t2 2

204 = 50

t=

u2 = 2500 + 204

t = t2 - t1 = 10 sec u = 52 m/s

EXAMPLE: 03 A car starts from rest moving along a line, first with acceleration

a= 2 m/s 2, then uniformly and

finally decelerating at the same rate and comes to rest. The total time of motion is 10 sec. The average speed during this time is 3.2 m/s. How long does the car move uniformly (A) 4 sec

(B) 6 sec

(C) 5 sec

(D) 3 sec

Solution: Let the car accelerate for time’t’ and move uniformly with v = at for time t1 since the magnitudes of acceleration and deceleration are same, the time of deceleration is also 't'. t + t 1 + t = 10 sec

Average speed =

Distance = time

1 2 at 2

at t 1 10

1 2 at 2

= 3.2

10 t 1 2 2

2

2t + 2tt1 = 32

2

2

10 t 1 t1 2

32

Solving t1 = 6 sec This problem can be solved using velocity time graph also.

EXAMPLE: 04 A particle has an initial velocity of 3 i

4 j m/s and a constant acceleration

4 i 3 j m/s2. Its

speed after 1 sec will be equal to (A) zero

(C) 5 2 m/s

(B) 10 m/s

(D) 25 m/s

Solution:

V

u a t = 3i

4j

4i

3 j` (1) 72

Speed = magnitude of V =

=7 i - j

1 2 = 5 2 m/s

EXAMPLE: 05 An aeroplane flies along a straight line from A to B with air speed V and back

u

B

again with the same air speed. If the distance between A and B is l and a steady wind blows perpendicular to AB with speed u, the total time taken for the round trip is (A)

(C)

2 V

2

(B)

2 V

(D)

2 2

V u

VR

V

V2

A

u2

2 V2

u2

Solution:

B

u

The resultant velocity of the plane must be along AB during forward journey.

 t1 = VR

VR

 V2

V

u2

A During return journey, the resultant velocity of the plane must be along BA t2 =

 VR

 V2

Total time t = t1 + t2 =

u2

2 V2

u2

EXAMPLE: 06 A particle is thrown with a speed 'u' at an angle angle

with the horizontal. When the particle makes an

with the horizontal its speed changes to v. Then

(A) v = u cos

(B) v = ucos cos

(C) v = u cos sec

(D) v = usec cos

Solution: Since the horizontal component of the velocity of a projectile always remains constant u cos =v cos

v=ucos sec

EXAMPLE: 07 Two shells are fired from cannon with same speed at angle

and

respectively with the horizontal.

The time interval between the shots is T. They collide in mid air after time’t’ from the first shot. Which of the following conditions must be satisfied. (A)

>

(B) t cos = (t -T) cos

(C) (t-T) cos =cos

(D) (usin )t -

1 2 1 gt =(usin ) (t-T)- g(t-T)2 2 2

Solution: When they collide, their 'x' and 'y' components must be same ucos t = u cos (t-T) (usin ) t -

cos t = cos (t-T)

1 2 1 gt = (usin ) (t-T) g (t-T)2 2 2

Since cos = cos

1

cos

>

< cos

and

T t

and T < t

EXAMPLE: 08 A particle is projected from a point 'p' with velocity 5 2 m/s perpendicular to the surface hollow right angle cone whose axis is vertical. It collides at point Q normally on the inner surface. The time of flight of the particle is (A) 1 sec

(B)

2 sec

(C) 2 2 sec

Solution: It can be seen from the diagram that V becomes perpendicular to u .

u = ucos450 i + u sin45 j

(D) 2 sec

V = u + a t = (ucos 45 i + usin45 j ) - (gt) j When V becomes perpendicular to u , V . u = 0 u2 cos2 45 + u2 sin2 45 - (usin45) g t = 0

t=

u = 1 sec g sin 45

EXAMPLE: 09 A man walking Eastward at 5 m/s observes that the wind is blowing from the North. On doubling his speed eastward he observes that the wind is blowing from North East. The velocity of the wind is (A) (5i+5j) m/s

(B) (5i - 5j) m/s

(C) (-5i +5j) m/s

(D) (-5i - 5j) m/s

Solution: Let V w

V1 i

V2 j

In the first case V wm

Vw

V m = V1 i

Since no component along East is observed V1 - 5 = 0

V2 j - 5 i V1 = 5 m/s

In the second case

V wm

Vw

V m = (V1 i + V2 j ) - (10 i )= V1

10 i

V2 j

Since the wind is observed from North East the components along North and East must be same V1 - 10 = V2

V2 = - 5 m/s

V w = (5i - 5j) m/s EXAMPLE: 10 From a lift moving upward with uniform acceleration 'a', a man throws a ball vertically upwards with a velocity V relative to the lift. The time after which it comes back to the man is (A)

2V g a

(B)

V g

a

(C)

2V g a

(D)

2 Vg g a2 2

Solution: Since the velocity of the ball is given relative to the lift V bl = V j When the ball comes back to the man, its displacement relative to the lift is zero S bl = 0

a bl = a b

a l = (-g) j - a j = - (g + a) j

Applying S = ut + 1/2 at2 in relative form

S bl

V bl t +

0 = Vt j +

1 2

1 a bl t2 2 g a j t2

t=

2V g a