Motion in One Dimension

Motion in one Dimension KINEMATICS AND DYN AMICS The part of mechanics which deals with the description of the motion of an object without considering reason of the origin is called kinematics. Where as, the study of the motion of an object related to its cause is called dynamics. (i) Motion in one dimension: Motion of an object in a straight line is called one dimensional (1-D) motion. The position of a particle in one dimensional motion can be described by only one variable (say x). For a particle moving along a straight line (1-D motion) all the vector quantities such as position, velocity, displacement and acceleration have only one non-zero component. (ii) Motion in two dimension: Motion of an object in a plane is called two dimensional (2-D) motions. For 2-D motion velocity or acceleration can be described by two components in any two mutually perpendicular directions in Cartesian coordinate system i.e. its position, velocity, displacement and acceleration can have two nonzero components. (iii) Distance & Displacement: Displacement: - The change in position of a body in a particular direction is known as displacement. It is a vector quantity and its unit in SI system is meter. The shortest distance between the initial and final positions of the object in a specified direction. Distance: The total length of actual path traversed by a body in a certain interval of time is called distance. It is the actual path travelled by an object between its initial and final positions. It is a scalar quantity and its unit in SI is meter. Displacement may be positive, negative or zero but distance is always positive. If a particle moves in a straight line without change in its direction, the magnitude of displacement is equal to the distance travelled. Otherwise it is always less than it. Thus, Displacement  distance Example 1 : What will be the distance and displacement while moving in a circle from A to B and then B to A as shown in adjoining figure? A R

Solution: Physical quantity and direction

Half cycle AB or BA

R

B

Full cycle AA via B

R 2R 1. AB, when particle moves from A to B. 2. BA, when particle moves from B to A.

Distance Displacement Direction of displacement

o

2R 0

(iv) Average Speed and Velocity: The average speed of a particle in a given interval of time is defined as the ratio of the distance travelled to the time taken while, average velocity is defined as the ratio of the displacement to the time taken. If a particle moves from A to C through a path ABC. Then distance

 Δs 

travelled is

B

Y

 Δr

the actual path length ABC, while the displacement is,

   Δ r = rC - r A

Thus, if the distance travelled is

 Δs 

and displacement of a particle is

given time interval Δt then

A t ru st e d n am e f o r Qu a lit y E d uc at i o n… …

 Δr in a

C

A  rA

 rc

X

Chapter - 1 Motion in One Dimension

  Δs Δr v av = A ve ra g e spe e d = an d V a v = A ve ra g e ve lo city = Δt Δt (v) Instantaneous Speed and Velocity: Instantaneous speed and velocity are defined at a particular instant and are

   Δs ds Δr dr given by , v = lim = and V = lim = Δt  0 Δt Δt  0 Δt dt dt

(vi) Average and Instantaneous Acceleration: Average acceleration is defined as the change in velocity interval t .

 V over a time

 V aav  t 

Hence,

  ΔV dV  lim  Δt  0 Δt dt

The instantaneous acceleration of a particle is the rate at which its velocity is changing at that instant i.e., 

ains Example 2 :

A particle moves along a semi circle path A to B in a time T as shown in the following fig. (a) Determine the average speed of the particle. (b) Determine the average velocity of the particle.

distance R (a) The average speed of the particle =  time T displacement 2R (b) The average velocity of the particle=  time T

Solution:

A

o

B

EQUATIONS OF MOTION EQUATIONS OF MOTION FOR UNIFORM ACCELERATION Consider a particle moving along a straight line with constant acceleration a. Let u be the initial velocity of the particle and v be the velocity at time t. By definition,

a

dv  dt

Integrating,

dv = a dt v

t

t

u

0

0

 dv   a dt  a dt 

(a being a constant can be taken out of the integral) v = u + at 

dx  Further, v = dt But,

v = u + at,

…(i)

dx = vdt



dx = (u + at)dt s

t

0 t

0

t

0

0

Integrating within appropriate limits, we get,



v – u = at

 dx   (u  at)dt

s  0  u dt  a t dt



s  u(t  0) 



s  ut 



1 a t 2  02 2



  

t  t 2 02  t 2  dt t and  0 0 t dt   2  2   2  t

1 2 at 2

Again, going back to definition of a, we have,

…(ii)

a

dv dt

Multiply and divide RHS of above equation by dx, we get,

 dx   dv  a    dt   dx 

Chapter - 1 Motion in One Dimension

 av

dv dx



adv = adx



s

u

0

 v dv  a dx

Integrating within appropriate limits, we get, 2

v

2

v u   a(s  0)  2 2



v 2  u2  2as

v2 2

s

v

 ax u

…(iii)

 u  v s t  2 

From (i) and (iii) we also conclude that,

0

…(iv)

Equation (iv) is a special equation, as we do not have a in this equation, but still it belongs to motion, which is uniformly accelerated. To conclude, we again write the equation of motion for constant acceleration below. v = u + at ;

uv  s t  2 

1 2 2 at ; v  u2  2as ; 2

s = ut +

EQUATION OF MOTION FOR VARIABLE ACCELERATION Case I when acceleration a of the particle is a function of time i.e.

dv  f(t) dt



A = f(t)



dv = f(t)dt

Integrating both sides within suitable limits, we get, v

t

 dv   f(t)dt u

Case II





0



v=u+

0

dv = f(x) dt

 v

dv = f(x) dx



 f(t)dt 0



dv dx = f(x)  dx dt

vdv = f(x)dx

v

s

u

0

 vdv   f(x)dx 

Integrating we get,

s

v 2 u2   f(x)dx  2 2 0

s

v 2  u2  2 f(x)dx 0

When acceleration a of the particle is a function of velocity i.e.,

dv  f(v) dt



a = f(v)

Option 1

Option 2

dv dt  f(v) Integrating,

vdv  dv vdv   f(v)  dx   a   dx f(v)  dx  x v v vdv vdv Integrating,  dx    x  x0   f(v) f(v) x0 u u

v

t

v

0

u

 dt  

dv f(v)

v

t u

th

dv f(v)

The displacement by the body in n second is given by , Example 3:

t



When acceleration a of the particle is a function of distance i.e. a = f(x)

Case III

t

v – u  f(t)dt

a sn  u  (2 n  1) 2

Choose the correct alternative(s) The displacement of a particle moving along the x axis is given by x = a(t-1) + b(t-1) Where a and b are constants; x is measured in meters and t is in second. The motion start at t = 0. (A) The initial velocity of the particle is a (B) The acceleration of the particle is 2b (C) The initial velocity of the particle is (a + b)

Chapter - 1 Motion in One Dimension Solution:

(D) The particle starts its motion from the origin at t = 1sec. x = a(t -1) + b(t -1) …(1)

dx = a + b , so the initial velocity of the particle is (a + b) dt putting t = 1sec.  x = 0 i.e the particle starts its motion from the origin at t = 1sec.Therefore C & D are the correct answers. Example 4:

Solution:

A ball thrown up from the ground reaches a maximum height of 20m. Find (A) Its initial velocity (B) the time taken to reach the highest point (C) the time at which it is 15m above the ground (D) its velocity just before hitting the ground (E) its displacement between 0.5 sec and 2.5 sec. 2 2 (A) Using v = u +2as for upward motion 2 0 = u + 2(-9.8)  (20)  u= 19.8 m/s (B) t = (C)

(D)

(E)

v -u 0  19.8   2.02 sec 9.8 a

using s = ut + ½gt 15 = 19.8t - ½ 9.8 t  t= 1.01 sec & 3.303sec. There are two solutions because At t = 1.01 ball is going up and at t= 3.303sec it is coming down. displacement is zero for the complete up-down trip. 2 2 2 2 v = u + 2 a (0) (by using v = u + 2as) 2 2  v =u  v = u = –19.8 m/s 2 height at t = 0.5 sec. can be calculate using s = ut + ½ at 2 y1 = 19.8 (0.5) – 4.9 (0.5) = 8.68 m 2 height at t = 2.53 is, y2 = 19.8 (2.5) – 4.9 (2.5) = 18.9 m displacement = y2 – y1 = 18.9 m – 8.68 m = 10.2 m 2

2

Common Error in kinematics

S = ut +

In using kinematical eqn. (1D),

1 2 at 2

Here 'S' in the displacement i.e. s = xf - xi. Its common error for the student to think that displacement and distance are the same. Example 5: Solution:

A ball is projected with a velocity of 20 m/s vertically. Find the distance travelled in first three 2 second. (use g = 10m/sec ) Here the problem is to find the distance. We can calculate that the direction of ball is changes at t = 2s. (From v = u + at, since v = 0 at highest point therefore 0=20 - 10t  t = 2s) Distance travelled in first two seconds ( Distance = Displacement, because velocity does not change direction in one dimension)

 S1 = 20× 2 -

1 ×10 × 4 = 40 – 20 = 20 m (upward) 2

Distance travelled in next one second ,

1 S2 = - ×10 ×1 = -5m downward 2 So total distance travelled by the ball in first three seconds = 20 +5 = 25m Example 6: Solution:

2

A particle moves along the x-axis according to x = 4t – t . Find the distance travelled from t = 0s to t = 3s. The direction of velocity will change when V = 0 at t = 2 sec. 3



Distance = vdt = 0

Example 7:

Solution:

1

2

3

0

1

2

 (4  2t )dt   (4  2t )dt   (4  2t )dt = 3+1+ 1 =5 meter

A particle moving with uniform acceleration from A to B along a straight line has velocities v1 and v2 at A and B respectively. If C A v1 is the mid point between A and B then determine the velocity of the particle at C. Let distance between A and B be x and V be the velocity of the particle at C, then

C

B v2

Chapter - 1 Motion in One Dimension 2

2

V = v1 + 2a (x/2)

 v 2  v1   t   

Where a is the acceleration of the particle; i.e.

t is the time taken by the particle to travel from A to B and x is the total distance between A and B,

 v1  v 2  t  2 

i.e. x 

 v1  v 2   v 2  v 1   2  t  t  Example 8:

Solution:

Thus, V –v1  2

2

v12  v 22 2

A particle travels according to the equation a  A  Bv where a is the acceleration. A and B are constants, v is the velocity of the particle. Find its velocity as a function of time. Also find terminal velocity. V

dv  A  B dt or

loge

A  B   Bt A

t

dv 0 A  B  0 dt

or

 

A  B  Ae Bt or

or

 1 loge  A  B 0  t B





A 1 e Bt B



At terminal velocity, acceleration = 0, that is, t  A / B

X–T, V–T & A–T GRAPHS FOR MOTION IN ONE DIMENSION (i) Variation of displacement (x), velocity (v) and acceleration (a) with respect to time for different types of motion. Displacement(x) a. At rest

Velocity(v)

Acceleration (a) a

v

x x=constant

t

O

b. Motion with constant velocity

x

v0 O

c. Motion with constant acceleration

x

t

O

t

O a

v v = v0+a0t

a = constant a0

O

d. Motion with constant deceleration

v = constant

t

x = v0t +(1/2)a0t2

t

O

a

v x = x0+ v0t +x0t2

x0

t

O

t

x

v0 O

t

a

v x = v0t -(1/2)a0t2

a = constant

v0 O

O

t

O

t

O

t

t

a0

(ii) Displacement calculation from Velocity - Time Graphs: The displacement during an interval between time ti and tf is the area bounded by the velocity curve and the two vertical lines t = ti and t = tf, as shown in figures (a) and (b).

Chapter - 1 Motion in One Dimension v v4 v3 v2 v1 O ti

t1

t 2

t3

t

t 4

tf

(a) For each segment of motion, the velocity is constant. The displacement  xi during the i th interval is the area vi  ti. So the total displacement is  x  x = v1  t1 + v2  t2 + v3  t3 + v4  t4 x  v i t i







(b) When v vs. t graph is a smooth complex curve. Area bounded by the curve and time axis between t=ti and t=tf is the displacement. The area under the curve may be obtained by using integration.

(iii) Velocity calculation from Acceleration - time Graphs: Given an acceleration–versus–time graph, the change in velocity between t = ti and t = tf is the area bounded by the acceleration curve and the vertical lines t = ti and t = tf a a0

O

t

tf

ti

(b) When a vs t graph is a complex curve, the area (a) When the area under the a vs t is a smooth under the curve may be obtained by using integration curve then the change in velocity v  a0 t The velocity-time graph of a particle moving along a straight line is shown in following figure. (i) If the particle starts its motion from x = –4m, then draw the (a–t) and (x–t) graphs. (ii) Find the displacement of the particle at t = 3 s

v

velocity

Example 9:

4 m/s O

Solution:

(i) x

a

+4 m/s2 +2

O

1

2

3

4

t(s)

x(m)

-2

2

O

4

t(s)

-4

(ii) x  x0  [Area of v-t graph]tt 30   1  4  2 x    4  (2)(4)   (3  2)   2  2  

Example 10:

or

x  3m

From the given  a  t  graph, draw  v  t  and  s  t  curves.

2 Time

4

t

Chapter - 1 Motion in One Dimension a(ms-2) +10 0

10

15

t(s)

5 -10

Solution:

Velocity is the area under  a  t  curve, thus at the end of 5s, velocity v  50ms 1. Since there is no acceleration in the interval (5 -10 s), therefore velocity remains constant, that is, 50ms 1. In the interval (10 – 15 s) the particle is retarded (Area = -50). Thus, we get the (v – t) curve as shown in figure (a) Hence particle comes to rest.

v  t 

To draw  s  t  graph [Fig (b)]. We take help of  v  t 

v  t 

graph. Area under

s (m )

500

curve is the distance

375

travelled. 50  5  125m 2 Distance moved in  5  10 s  is 5  50  250m Distance moved in  0  5s  is

125

50  5 Distance moved in 10  15s  is  125m 2

5 10 15 (s - t) c u r v e

0

t

Example 11:

A ball drops from a height 19.6 m above the ground, it rebounds and rises to the same height. Plot  v  t  and  x  t  graphs

Solution:

1 2 gt Þ 19.6 ´ 2 = 9.8 t 2 or t ¾ 2s 2 This is the time for the ball to come to the ground. x=

v(ms-1)

x(m)

19.6

19.6

t

0 1

2

3

4

-19.6

t

0 1

2

3

4

Velocity – Time graph Distance –Time graph

Solution:

To plot the

and



v  t 



graph for the given

graph we



t a



Plot

t x

t v



Example 12:



graph.

can see from the (a-t)

a(ms-2)

graph, that a  2 t 0  t   5 s  v 



5 0

adt 

v  u  at  7 5m s



5 0

5 s 1

2t2 2 td t  2

10

5

 2 5m s

1

0

 t  10s  25  10  5

a t th e e n d o f 1 0 s

0 0

5

10

t

Chapter - 1 Motion in One Dimension

t3 For  x  t  graph, x = ò v.dt = 0 3

5

5

= 0

125 m (0 < t < 5s ) 3

x = area under  v  t  curve (5s < t < 10 s) = 250 m. Example 13:

A particle starts from rest with acceleration  for some time and after achieving a maximum velocity starts retarding at rate  and finally comes to rest. If total time taken is t then Determine (a) maximum velocity (b) total distance travelled. (A) (C)

Solution:

 t  t 2 ,          t2 t ,     2 2

(B)

 t  t 2 ,    2    

(D)

    t ,    

t2 2

(B) Let the particle accelerate for a time t then maximum, velocity v   t1 v





t t1

t-t1

Since it retards at a rate  and finally comes to rest therefore 0   t1    t  t1  Or t1 

t  t  vmax     

Distance travelled = area under (v -t) graph

1 vmax.  total time  2 1  t  t 2   t= 2  2     

Example 14:

Solution:

The displacement of a particle moving in one dimension is given by t  x  3 where x is in meter and t in second. The displacement, when the velocity is zero is (A) 3 m (B) 1 m (C) 1.8 m (D) None of these dx 2 (D) x = (t - 3) and = 2 (t - 3) dt dx v= = 0 when t = 3 dt 2

x = (3 - 3) = 0

Example 15:

A body is moved along a straight line path by a machine delivering constant power. The distance moved by the body in time t is proportional to

Chapter - 1 Motion in One Dimension

Solution:

3

 dx 

t

2

1

t 4 (B) (C) dV P.dt Power P  F.v  m v  vdv   dt m v2 P or = t + C when t = 0, v = 0 \ C=0 2 m 2Pt dx 2Pt or v = or = m dt m

(A)

3 2P 12 t dt or x  t 2 ,  m

t

1

2

So option (a) is correct.

(D)

t

3

4

Chapter - 1 Motion in One Dimension

ASSIGNMENT LEVEL - I 1.

2.

3.

4.

du (t) = 6 - 3u (t) dt –1 Where (t ) at time t is in ms and t is in seconds. If the body was at rest at t = 0, test the correctness of the following results (A) The terminal speed is 2 ms –1. (B) The magnitude of the initial acceleration is 8 ms2. (C) The speed varies with time as u (t) = 2(1 - e - 3t ) (D) The speed variation is 2 ms–1 when the acceleration is half the initial value. A driver applies brakes to the vehicle on seeing traffic signal 400 m ahead. At the time of applying the brakes vehicle was moving with 15 ms-1 and then starts retarding with 0.3 ms-2. The distance of vehicle after 1 min from the traffic light: (A) 25m (B) 375m (C) 360m (D) 40m A particle is moving along a circular path of radius 5m and with uniform speed 5 m/s. What will be the average acceleration when the particle completes half revolution? (C) 10 m/s2 (D) 10/m/s2 (A) zero (B) 10 m/s2 Which of the following graph correctly represents velocity-time relationship for a particle released from rest to fall freely under gravity? v v v v The motion of a body is given by the equation

t

t

5.

6.

7.

8.

9.

10.

12.

t

(A) (B) (C) (D) A bus accelerate uniformly from rest and acquires a speed of 36 km/hour in 10 seconds. The acceleration is (A) 1000 m/sec2 (B) 1 m/sec2 (C) 100 m/sec2 (D) 10 m/sec2 A stone is thrown vertically up from the ground. It reaches a maximum height of 50 meters in 10 sec. After what time it will reach the ground from the maximum height position? (A) 1.2 sec (B) 5 sec (C) 10 sec (D) 25 sec A body thrown up with a velocity reaches a maximum height of 50 meters. Another body with double the mass thrown up, with double the initial velocity will reach a maximum height of (A) 100 m (B) 200 m (C) 10 m (D) 400 m The reaction time for a car driver is 0.9 sec. If the car travelling initially with 36 Km/hr is stopped by the driver in two seconds after observing a signal by the deceleration of 5 m/s2, the total distance traveled by the car before coming to rest is (A) 19 m (B) 9 m (C) 10 m (D) 28 m A car starts from rest. Attains a velocity of 36 km/h with an acceleration of 0.2 m/s2, travels 9 km with this uniform velocity and then comes to halt with a uniform deceleration of 0.1 m/s2. The total time of travel of the car is (A) 1050 s (B) 1000 s (C) 950 s (D) 900 s A car travelling on a straight track moves with uniform velocity V1 for some time and with uniform velocity V2 for next equal time, the average velocity is given by (A)

11.

t

V1V2

VV (B) 1 2 2

 1 1 (C)    V1 V2 

1

 1 1 (D) 2    V1 V2 

1

A lift is moving with a retardation of 5 m/s2. The percentage change in the weight of person in the lift is (g = 10 m/s2) (A) 100 (B) 25 (C) 50 (D) 75 The distances traveled by a body starting from rest and travelling with uniform acceleration, in successive intervals of time of equal duration will be in the ratio

Chapter - 1 Motion in One Dimension

13.

14.

15.

16.

17.

18.

19.

20.

(A) 1 : 2 : 3 (B) 1 : 2 : 4 (C) 1 : 3 : 5 (D) 1 : 5 : 9 The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 m is 0.34 m/s–1. If the change in velocity of the body is 0.18 ms–1 during this time its uniform acceleration is. (A) 0.01 m/s2 (B) 0.02 m/s2 (C) 0.03 m/s2 (D) 0.04 m/s2 2 Equation of position (x) with time (t) is given by equation x = 3t + 7t2 + 5t + 8m. The acceleration at time t = 1 sec. is : (A) 18 m/sec2 (B) 32 m/sec2 (C) Zero (D) 14 m/sec2 A man slides down an inclined plane and drops a bag from the position to the ground. If velocities of man and bag on reaching the ground are v M and v B respectively then: (A) v M  v B (B) v M  vB (C) v M  v B (D) depend on weight One car moving on a straight road covers one third of the distance with 20 km/hr and the rest with 60 km/hr. The Average speed is 2 (D) 36 km/hr (A) 40 km/hr (B) 80 km/hr (C) 46 km / hr 3 A particle covers 50m. distance with 40 kmph and rest half distance with 60 kmpl then the average speed of car is : (A) 100 m (B) 150 m (C) 200 m (D) 250 m A car travels half distance with 40 kmph and rest half distance with 60 kmph then the average speed of car is : (A) 40 kmph (B) 48 kmph (C) 52 kmph (D) 60 kmph A particle is moving east-wards with a velocity of 15 m/s. In a time of 10 seconds, the velocity changes to 15 m/s north-wards. Average acceleration during this time is, in m/s2 3 3 (A) north-east (B) 3 2 north-east (C) north-west (D) 3 2 north-west 2 2 A car starts from rest and travels with uniform acceleration  for some time and then with uniform retardation  and comes to rest. If the total travel time of the car is 't', the maximum velocity attained by it is given by  1   1  (A) (B) (C) (D) .t .t 2 .t .t 2 (   ) 2 (   ) (   ) 2 (   )

Chapter - 1 Motion in One Dimension

LEVEL - II 1.

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Water drops fall from a tap on to the floor 5.0 m below at regular intervals of time, the first drop striking the floor when the fifth drop begins to fall. The height at which the third drop will be from ground, at that instant when first drop strikes the ground, will be, taking g = 10.0 m/sec2. (A) 1.25 m (B) 2.15 m (C) 2.75 m (D) 3.75 m A stone is dropped freely, while another is thrown vertically downward with an initial velocity of 2 ms2 from the same point, simultaneously. The time required by them to have a distance of separation 22 m between them is (A) 11 sec. (B) 5.5 sec. (C) 44 sec. (D) 22 sec. A body is thrown vertically upwards with an initial velocity u reaches maximum height in 6 seconds. The ratio of distance traveled by the body in the first, second and seventh second is : (A) 1 : 1 (B) 11 : 1 (C) 1 : 2 (D) 1 : 11 A particle travels 10 m in first 5 sec and 10 m in next 3 sec. assuming constant acceleration what is the distance traveled in next 2 sec. (A) 8.3 m (B) 9.3 m (C) 10.3 m (D) None Acceleration of a particle changes when : (A) Direction of velocity changes (B) Magnitude of velocity changes (C) both of above (D) speed changes The relation 3t  3x  6 describes the displacement of a particle in one direction where x is in meters and t in sec. The displacement, when velocity is zero, is (A) 24 m (B) 12 m (C) 5 m (D) zero The displacement 'x' of a particle moving along a straight line at time t is given by x  a0  a1t  a2 t 2 what is the acceleration of the particle? (A) a1 (B) a2 (C) 2a2 (D) 3a2 A body starts from rest and has an acceleration 20 cm/sec2. What is the distance covered by the body in first 8 sec ? (A) 160 cm (B) 640 cm (C) 1280 cm (D) 1640 cm Pick up the correct statements: (A) Area under a-t graph gives velocity (B) Area under a-t graph gives change in velocity (C) Path of projectile as seen by another projectile is a parabola, (D) A body, whatever be its motion, is always at rest in a frame of reference fixed to the body itself. A body is moving in a circle at a uniform speed  . What is the magnitude of the change in velocity when the radius vector describes an angle :   (A)  cos  (B) 2 cos   (C)  sin  (D) 2 sin    2  2 A bicyclist encounter a series of hills uphill whose speed is always v1 and down hill speed is always v2. The total distance travelled is , with uphill and downhill portions of equal length. The cyclist's average speed is: v 2  v 22 v v v v 2 v1 v 2 (A) 1 2 (B) 1 (C) 1 2 (D) 2 v1  v 2 v1  v 2 v1  v 2 Choose the wrong statement: (A) Zero velocity of a particle does not necessarily mean that its acceleration is zero. (B) Zero acceleration of a particle does not necessarily mean that its velocity is zero. (C) If the speed of a particle is constant, its acceleration must be zero. (D) None of these Displacement (x) of a particle is related to time (t) as x = at + bt2 – ct3 where a, b and c are constants of motion. The velocity of the particle when its acceleration is zero is given by: b2 b2 b2 b2 (A) a  (B) a  (C) a  (D) a  c 2c 3c 4c

Chapter - 1 Motion in One Dimension

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A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is (A) 4.0 m/s (B) 5.0 m/s (C) 5.5 m/s (D) 4.8 m/s The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v0. The distance travelled by the particle in time t will be 1 1 1 1 (A) v 0 t + bt 2 (B) v 0 t + bt 3 (C) v 0 t + bt 3 (D) v 0 t + bt 2 3 3 6 2 dv(t) The motion of a body is given by the equation = 6.0 - 3v(t) . where v(t) is speed in m/s and t in dt sec. If body was at rest at t = 0 (A) The terminal speed is 2.0 m/s (B) The speed varies with the time as v(t) = 2(1 – e–3t) m/s (C) The speed is 0.1 m/s when the acceleration is half the initial value (D) The magnitude of the initial acceleration is 6.0 m/s2 A particle starts from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be (A) 110 m/s (B) 55 m/s (C) 550 m/s (D) 660 m/s If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is 1 1 (B) ut - gt 2 (C) (u – gt)t (D) ut d (A) gt 2 2 2 A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that (A) Velocity is constant (B) Acceleration is constant (C) Kinetic energy is variable (D) It moves in a circular path The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance traveled by the particle in four second is (A) 60 m (B) 55 m (C) 25 m (D) 30 m