8. Motion Class 9 NCERT Exemplar Solutions Chapter 8 (CBSE Science)
CBSE Class 9 Science Chapter 8 : Motion – NCERT Exemplar Questions with Answers
Chapter 8 : Motion
Multiple Choice Questions
1. A particle is moving in a circular path of radius r. The displacement after half a circle would be:
(a) Zero (b) πr (c) 2r (d) 2πr
Answer: (c) 2r
[ Displacement is the straight-line distance from start to end point. After half a circle, the particle reaches the diametrically opposite point, so displacement = diameter = 2r ]
2. A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) u/g (b) u²/2g (c) u²/g (d) u/2g
Answer: (b) u²/2g
[ When a body is thrown vertically upward with initial velocity u, at the greatest height its final velocity becomes 0.
Here, v = 0 , a = – g , s = h
We have, v² = u² + 2as
0 = u² – 2gh ⇒ 2gh = u² ⇒ h = u²/2g
So, the greatest height is u²/2g. ]
3. The numerical ratio of displacement to distance for a moving object is
(a) always less than 1 (b) always equal to 1 (c) always more than 1 (d) equal or less than 1
Answer: (d) equal or less than 1
[ Displacement ≤ distance . They are equal only for straight-line motion in one direction. Hence ratio ≤ 1. ]
4. If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity (b) uniform acceleration (c) increasing acceleration (d) decreasing acceleration
Answer: (b) uniform acceleration
[ Because when displacement is proportional to t² , the object moves with uniform acceleration.]
5. From the given v – t graph (Fig. 8.1), it can be inferred that the object is
Fig. 8.1
(a) in uniform motion (b) at rest (c) in non-uniform motion (d) moving with uniform acceleration
Answer: (a) in uniform motion
[ The v–t graph is a horizontal straight line (constant velocity), indicating the object moves at constant speed with zero acceleration—uniform motion.]
6. Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m/s . It implies that the boy is
(a) at rest (b) moving with no acceleration (c) in accelerated motion (d) moving with uniform velocity
Answer: (c) in accelerated motion
[ Even though the merry-go-round moves at constant speed (10 m/s), the boy moves in a circle. Circular motion requires centripetal acceleration towards the center, so he is in accelerated motion. ]
7. Area under a v – t graph represents a physical quantity which has the unit
(a) m² (b) m (c) m³ (d) m/s
Answer: (b) m
[ The area under a velocity–time (v–t) graph gives displacement, and the SI unit of displacement is metre (m).]
Area = velocity × time = (m/s) × s = m (metre) ]
8. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in Fig. 8.2. Choose the correct statement.
Fig. 8.2
(a) Car A is faster than car D. (b) Car B is the slowest. (c) Car D is faster than car C. (d) Car C is the slowest.
Answer: (b) Car B is the slowest.
[On a distance – time graph, a flatter slope indicates a lower speed . Since line B has the least steep slope , Car B is the slowest .]
9. Which of the following figures (Fig. 8.3) represents uniform motion of a moving object correctly?
Fig. 8.3
Answer: Correct option (a)
[In uniform motion , an object covers equal distance in equal time intervals . This is represented by a straight line on a distance – time graph, showing constant speed. ]
10. Slope of a velocity – time graph gives
(a) the distance (b) the displacement (c) the acceleration (d) the speed
Answer: (c) the acceleration
[The slope of a velocity–time graph represents the rate of change of velocity with time, which is called acceleration.]
11. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on straight road (b) If the car is moving in circular path
(c) The pendulum is moving to and fro (d) The earth is revolving around the Sun
Answer: (a) If the car is moving on straight road
[ When motion is along a straight line in the same direction, the path length (distance) exactly equals the straight-line separation from start to end (displacement magnitude).]
Short Answer Questions
12. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify you answer.
Answer: No, the distance travelled by the object would not be zero. The displacement becomes zero when the object returns to its starting point. But distance is the total path covered.
For example, if a student walks 5 m forward and then 5 m back, displacement = 0, but distance travelled = 10 m.
So, distance can be non-zero even when displacement is zero.
13. How will the equations of motion for an object moving with a uniform velocity change?
Answer: If an object is moving with uniform velocity, i.e., Acceleration a = 0
(i) We have, 
Velocity remains constant.
(ii) We have, 
(iii) We have, 
When velocity is uniform (constant), then acceleration is zero and the equations of motion become much simpler.
14. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in Fig.8.4. Plot a velocity–time graph for the same.
Fig. 8.4
Answer:
15. A car starts from rest and moves along the x-axis with constant acceleration 5 m/s² for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?
Solution: Here, u = 0 , a = 5 m/s² , t = 8 s
Velocity after 8 s:
m/s
We have, 
Remaining time 
Total distance 
The car covers 320 metres in 12 seconds.
16. A motorcyclist drives from A to B with a uniform speed of 30 km/h and returns back with a speed of 20 km/h . Find its average speed.
Solution: Given, Speed from A to B 
and Speed from B to A 
The average speed 
The average speed for the complete journey is 24 km/h.
17. The velocity-time graph (Fig. 8.5) shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds.
Fig. 8.5
Answer:
18. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
Answer:
Long Answer Questions
19. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?
Solution: Given, object 1 height = 150 m and Object 2 height = 100 m
Initial velocities 
(dropped from rest)
Time t = 2s
We have, 
This distance is same for both objects because acceleration and time are same.
Initial height difference = 150 m – 100 m = 50 m.
Thus, difference remains 50 m.
20. An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start.
Solution: Given, u = 0 , 
,
We have, 
m/s² .
Again, 
m/s
Next 4 seconds (from 2 s to 6 s) :
Here, 
m/s ,
m/s²
Since acceleration remains constant at 10 m/s² throughout, Velocity after 7 s from start:
We have, 
m/s
The velocity after 7 seconds is 70 m/s.
21. Using following data, draw time - displacement graph for a moving object:
|
Time (s) |
0 |
2 |
4 |
6 |
8 |
10 |
12 |
14 |
16 |
|
Displacement (m) |
0 |
2 |
4 |
4 |
4 |
6 |
4 |
2 |
0 |
Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.
Answer:
22. An electron moving with a velocity of 
m/s enters into a uniform electric field and acquires a uniform acceleration of 
m/s² in the direction of its initial motion.
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time?
Solution: Given, 
m/s , 
m/s² and 
m/s
(i) We have,
(ii) We have, 
m
23. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.
Solution: We have, 
The distance travelled in 5 seconds, 
The distance travelled in 4 seconds, 
The distance between 4th and 5th second 
So, the distance travelled between 4th and 5th second is 
.
24. Two stones are thrown vertically upwards simultaneously with their initial velocities 
and 
respectively. Prove that the heights reached by them would be in the ratio of 
( Assume upward acceleration is –g and downward acceleration to be +g ).
Solution: Here, 
We have, 
Initial velocities 
and
and 
So, the heights reached are directly proportional to the squares of their initial velocities.
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