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Revision Question Bank
Motion
1.
A force of 10 N produces an acceleration of 2.5 ms–2, when applied on an object. What is
the mass of the object?
Solution:
Mass (m) =
2.
F
10N

 4kg
a 2.5ms2
A car moves with speed of 30 km / h for half an hour, 25 km /h for one hour and
40 km / h for 2 hours. Calculate the average speed of the car.
Solution:
Average speed =
Total distance
Total time
1
30   25  1  40  2 15  25  80
2
i.e., v av 

0.5  1  2
3.5
=
3.
120
 34.28 km / h
3.5
Satvik went on his bike from Delhi to Faridabad at 40 km/h and in the evening returned
back at speed of 60 km/h. What is his average speed for the entire journey? What is his
average velocity?
Solution:
Satvik went on his bike from Delhi to Faridaba) at 40 km/h and in the evening returned
back at speed of 60 km/h. What is his average sped for the entire journey? What is his
averad velocity?
4.
What is meant by uniform circular motion? Why do we need a force to keep a body
moving uniformly along a circular path? Name this force.
Solution:
A circular motion, in which the speed remains the same is called uniform circular motion.
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A force is required to keep the body in the circular path. In the
absence of the force, the body will tend to move
along its tangent (at any point). The force is named as centripetal force.
5.
Draw the shape of v-t graphs in the following cases
(a) Uniform retardation
(b) Non-uniform acceleration
Solution:
(a) Slope is negative (–ve) and constant, so uniform retardation.
(b) Slope is varying, non-uniform acceleration.
6.
Starting from a stationary position, Ajay paddles his bicycle to attain a velocity of 10 ms–1
in 25 s. Then, he applies brakes such that he again comes to rest after next 50 s. Calculate
the acceleration of the bicycle in both cases. Also find the total distance covered by Ajay.
Solution:
In first case, initial velocity u1 = 0, time t1 = 25 s and final velocity, v1= 10 ms–1
 acceleration, a1 =
v1  u1 10  0

 0.4ms2
t1
25
and distance covered in time t1 is
1
s1  u1t 1  a1t 12
2
1
2
= 0  2.5    0.4   25
2
= 0 + 125
= 125 m
After applying the brakes in second case,
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Initial velocity, u2 = 10ms –1
Time, t2 =50s
Final velocity,
v2 = 0
 acceleration,
a2 
v2  u2 0  10

 0.2ms2
t2
50
and distance covered in time t2 is
1
1
2
s2  u2t 2  a2t 22  10  50    0.2  50
2
2
= 500– 250 = 250 m
So, total distance covered by Anil =s = s1 +s2
=125+ 250 = 375m
7.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be
the distance covered and the displacement at the end of 2 min 20s?
Solution:
(i) Diameter of circular track = 200 m
 Radius of circular track =
200
= 100 m
2
Circumference of circular track
 2r  2 
22
4400
 100 
m
7
7
Total time given = 2 min 20 s
= 2×60+20 = 140s
According to question, athlete takes 40 s to complete one round.
 Distance covered in 40 s =
4400
7
m
 Distance covered in one round is equal to the circumference of the track).
Distance covered m 1 s =
4400
7  40
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or distance covered in 140 s =
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4400
 140
7  40
= 2200 m
= 2.2 km
(ii) As per the question, athlete completes one round in 40 s, i.e, after 40 s, athlete comes
back to its initial position or after 40 s his displacement is
zero. Similarly after 120s (40 ×3), his displacement is zero.
 Displacement after 2 min 20 s or 140 s
= Displacement after 20 s.
(
After 120 s (2 min), displacement is zero.)
In 20 s, athlete will cover half the circular track from A to B (in 40 s one complete round is
taken, so in 20 s half round will be taken).
Displacement after 20 s = AB|
(Initial position is A and after half round, final position is B)
= Diameter of circular track
= 200 m
8.
State which of the following situations are possible and give an example of each of these.
(a) a body moving with a constant acceleration but with zero velocity.
(b) a body moving horizontally with an acceleration in the vertical direction.
(c) a body moving with a constant velocity in an accelerated motion.
Solution:
(a) Possible, for example, a stone projected vertically has constant acceleration but has
zero velocity at the highest point.
(b) Possible, for example, a body projected at some angle will have horizontal motion,
while g acts vertically down.
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(c) Not possible. Since, in constant velocity, direction and magnitude are both to
be
constant. If constant speed is considered, then uniform circular motion is an example.
9.
The distance-time graph of two trains is given below. The trains start simultaneously in
the same direction.
(a) How much ahead of A is B when the motion starts?
(b) What is the speed of B?
(c) When and where will A catch B?
(d) What is the difference between the speeds of A and B
(e) Is the speed of both the trains uniform or non-uniform? Justify your answer.
Solution:
(a) B is ahead of A by 100 km.
(b)Speed of B=
QR 150  100
=
=25km/h
PR
20
(c)A will catch B at t = 2 h
when A has covered a distance of 150 km.
(d) Speed of A = Slope of line OA
=
OQ 150  0

 km / h  75km / h
OL
2
Speed of B= Slope of line PB
=
QR 50

 25km / h |
PR 2
Difference in speed of A and B = 75– 25
= 50 km/h
(e) Speed of both the trains is uniform as they cover equal distance in equal intervals of
time and s-t graph is a straight line for both trains.
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10. Derive the equations of motion for uniformly accelerated motion from velocity-time
graph.
Solution:
Equations of motion by graphical method Consider an object moving along a straight line
with an initial velocity u and uniform acceleration a. Suppose, it travels a distance, s in
time, t. As shown in figure its velocity-time graph
is a straight line. Here, OA= ED=u, OC = EB= v, OE = AD = t
(i) Equation for Velocity-time Relation
We know that, Acceleration = Change in Velocity/Time
or
a = BD/OE= (BE – ED) OE
or
a = (v — u) /t
or
v — u = at
or
v=u + at
This proves the first equation of motion.
(ii) Equation for Position-time Relation Distance travelled by an object in time t is s.
s = Area of the trapezium OABE + Area of QADE + Area of ADB
or
s=(OA×OE) +
1
×DB×AD
2
...(i)
Now, DB = BE – DE = v –u = at Putting this value for DB in Eq. (i), we get
1
s=ut + at2
2
This proves the second equation of motion.
(iii) Equation for Position-velocity Relation The distance travelled by an object in time t is
S = Area of the trapezium OABE = 1/2(EB + OA)×OE = 1/2(EB + ED)×OE
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Substituting EB, ED and OE with v, u and t, we get
s=1/2(v + u)t ...(ii)
But from the first equation of motion, we know that
v = u + at
or
t =(v– u)/a
Substituting t in Equation (ii) with this value, we get
s = (v + u)(v-u)/ 2a
= (v2 –u2)/2a or
v2 —u2 = 2as
v2 =u2 + 2as
This proves the third equation of motion.
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Chapter Test {Motion}
M:M: 30
1.
M: Time: 40 Min.
A train travels at a speed of 60 km/h for 0.5 h, 24 km/h, for the next 0.25 h and then at
72 km/h for the next 0.75 h. Calculate the total distance covered by the train and its
average speed.
[3]
Solution:
Distance covered by train in 0.5 h at a speed of 60 km/h
S1 =60×0.5 = 30 km
Distance covered by train in next 0.25 h at a speed of 24 km/h
S2 = 24 × 0.25 = 6km
and distance covered by train in last 0.75 h at a speed of 72 km/h.
S3 =72×0.75 = 54 km
Total distance covered by train
S=S1 +S2 +S3 =30+ 6+ 54 =90 km
Total time, t = t1, +t2 + t3
= 0.5 + 0.25 + 0.75 =1.5 h
 Average speed =
2.
Distance  s 
90km
=60 km/h

Time  t 
1.5h
Odometer measures displacement of the vehicle. Correct this statement.
[2]
Solution:
Odometer measures distance covered by a vehicle.
3.
(a) List two differences between speed and velocity
[4]
(b) When is a body said to have
(i) uniform velocity
(ii) variable velocity,
(c) How is the average velocity of a body calculated when its velocity changes at a
non-uniform rate?
Solution:
(a) Differences between speed and velocity are
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S.
No.
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Speed
Velocity
It is displacement covered per
unit time.
It has both magnitude and
(ii)
Speed has only magnitude.
direction.
(iii)
It is always positive
It can be positive and negative.
(b) (i) Velocity of an object is said to be uniform if it covers equal displacements in equal
(i)
It is the distance covered per
unit time.
intervals of time.
(ii) Velocity of an object is said to be non-uniform if it covers unequal displacements in
equal intervals of time or equal displacements in unequal intervals
of time or if its direction of motion changes.
(c) Average velocity is measured by the ratio of total displacement covered to the total
time.
If a body covers a total displacement s in a time t, it means its average velocity has a
magnitude,
v av 
4.
s
t
A train starting from rest, pick up a speed of 10 m/s in 100 s. It continuous to move at the
same speed for the rest 250 s. It is then brought to rest in the rest 50 s. Plot a speed-time
graph for the entire motion of the train.
Calculate:
(a) Acceleration of the train while accelerating.
(b) Retardation of the train while retarding
(c) The total distance covered by the train.
[3]
Solution:
The speed-time graph for the entire motion of the train is shown below
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(a) Since, starting from rest, the train picks up a speed of 10 ms–1 in 100 s, hence
acceleration of train while accelerating
a
10  0
 0.4 ms2
100
(b) Retardation of the train while retarding
a' 
0  10
 0.2ms2
50
(c) Total distance covered by the train
s = area OABC
= area ODA + area DABE + area EEC
=
1
1
×100 × 10 + 250 × 10 + × 50 ×10
2
2
= 500+2500 + 250 = 3250m
5.
or
3.25 km
The brakes applied to a car produce an acceleration of 6 m/s2 in the opposite direction to
the motion. If the car takes two seconds to stop after the application of brakes, calculate
the distance, it travels during the time.
[3]
Solution:
Here acceleration of the car, a = – 6 ms–1 (acceleration is being taken negative because its
direction is opposite to that of motion and
velocity is decreasing).
Time, t = 2s and final speed, v = 0
 From relation v = u + at the initial speed of the car is
u = v–at = 0–(–6)× 2 = +12ms–1
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Hence, total distance covered by the car in given time is given by
1
1
s=at + at2 =12 × 2 + ×(–6)×(2)2
2
2
= 24–12 = 12m
6.
State three equations of motion. Which of them describes
(a) Velocity-time relation
[2]
(b) Position-time relation
Solution:
Let an object starts linear motion with an initial velocity u and a uniform acceleration a.
Let after time t its final velocity becomes v and during this time it covers a total distance
s.
Then, the three equations of motion are
1
(ii) s=ut + at2
2
(i) v = u + at
(iii) v2 –u2 =2as
Out of these equations, the first equation (v = u + at) describes the velocity-time relation
and the second equation (s =ut +
7.
1 2
at ) describes the position-time relation.
2
Answer the following questions:
[3]
(a) Differentiate acceleration from velocity.
(b) Can a body have acceleration without change in magnitude of velocity?
Explain with an example.
(c) A motor boat starting from rest on a lake accelerates in a straight line at a constant
rate of 3m/s2 for 8 s. How far does the boat travel during this time?
Solution:
(a) Change of velocity per unit time is called acceleration. If during a time t, velocity of an
object changes uniformly from an initial value u to a final value v, then
Acceleration  a  

Change in velocity
time
vu
t
Velocity is the displacement of the body per unit time. It is expressed as
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velocity  v  
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Displacement d

time
t
(b) A particle can be accelerated even if its speed is constant because it means that only
the magnitude of velocity is constant and its direction of motion may change
and
consequently the velocity may change with time. In uniform circular motion although
speed of the moving particle remains constant, but the direction of
motion
continuously goes on changing with time, so as to compel the particle to move along a
circular path instead of a straight line path. Hence, uniform circular motion
is an
accelerated motion.
(c) Here initial velocity of motor boat u = 0, constant acceleration, a = 3ms–2 and time t
=8s.
 Total distance covered by the boat during the time
1
s  ut  at 2
2
1
2
= 0   3   8   96m
2
8.
A train accelerates uniformly from 36 km/h to 72 km/h in 20 s. Find the distance
travelled.
[2]
Solution:
Initial speed of train, u = 36 km/h = 10 ms–1
Final speed of train,
v=72km/h=
72  5 1
ms  20ms1
18
and time. T = 20 s
 Uniform acceleration,
a
v  u 20  10

 0.5ms2
t
20
From the equation, v2 – u2 =2as
The total distance travelled during given time is
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v2  u2 20  10 400  100
s=


 300
2a
2  0.5
1
2
9.
2
An object starts linear motion with a velocity u and under uniform acceleration a, it
acquires a velocity v in time t. Draw v-t graph. From this graph obtain all the equations of
motion.
[5]
Solution:
The v– t graph has been shown in figure. At t = 0, the initial velocity is u (at point A) and
then increases to V (at point B) in time t.
(a)Draw perpendicular lines BC and BE from point B on the time and velocity
axis
respectively. So that OA = u, OE = BC = v
Also, draw a line AD parallel to time axis so that OC = AD = t. Then change in velocity in
time t = BC - OA = BC – CD = BD but BC = v
and
CD = OA = u
Hence, BD =v–u
From the velocity-time graph, the acceleration of the object is given by
a
Change in velocity BD BD v  u



Time taken
AD OC
t
at = v — u
v = u + at
(b) According to the v–t graph the total distance covered by the object is obtained by the
area under the graph. Hence, distances travelled by the object = area of trapezium OABC.
area of rectangle OADC + area of  ADB = OA × OC +1/2 (AD × BD) but OA =ut OC = t, AD
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= OC = t and BD = change is velocity
= v–u=at
 s = ut +1 / 2 ( t x at) = ut +
1 2
at
2
10. A car moves with a speed of 30 kmh–1 for half an hour, 25 kmh–1 for one hour and
40 kmh–1 for two hours. Calculate the average speed of car.
[3]
Solution:
Ans. Distance travelled by the car in time
1
t1  h
2
with a speed V1=30kmh–1 is
1
s1  v1t 1  30   15km
2
Distance travelled by the car in time t2 = 1 h with a speed v2 = 25 kmh–1 is
S2 =v2rt2 =25 ×1 =25 km
and the distance travelled by the car in time t3 = 2 h with a speed v3 =40 kmh–1 is
s3 =v3t3 =40×2 = 80 km
 Average speed of car, is vav
=
total distance travelled
total time taken
=
s1  s2  s3 15  25  80 km

t1  t2  t3
 1 / 2  1  2 h
=
120km
 34.3kmh1
35h
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