NCERT Exemplar Chapter 8 Motion Class 9 Science Solutions

NCERT Exemplar Class 9 Science Chapter 8 Motion Solutions

NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion covers all the important questions and answers as well as advanced level questions. It helps in learning about the concepts of motion, Straight-line motion, Measuring the rate of motion, Uniform motion, Non-uniform motion, distance, displacement, average speed and velocity.

The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like rate of change of velocity like acceleration and retardation, graphical representation of motion, distance time graphs, velocity time graphs, uniform circular motion and velocity-time graph.

Chapter Name	Chapter 8 Motion
Book Title	NCERT Exemplar for Class 9 Science
Related Study	NCERT Solutions for Class 9 Science Chapter 8 Motion Revision Notes for Class 9 Science Chapter 8 Motion MCQ for Class 9 Science Chapter 8 Motion Important Questions for Class 9 Science Chapter 8 Motion
Topics Covered	MCQ Short Answers Questions Long Answers Questions

NCERT Exemplar Solutions for Chapter 8 Motion Class 9 Science

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

Solution

(c) 2r

After half revolution
Displacement = Final position-Initial Position
Diameter of the circle = 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

Solution

(b) u²/2g

v² = u² + 2as
Given,
v = 0, a = -g, s = H
Therefore,
0 = u^² -2gH
⇒ H = u^²/2g

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

Solution

(d) Equal or less than 1
Shortest distance between initial and end point is called displacement. Distance is the total path length. Displacement is vector and it may be positive or negative.

Distance is scalar and it can never be negative. The distance travelled by a body can be equal to or greater than the displacement which means ratio of displacement to distance is always equal to or less than 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

Solution

(b) Uniform acceleration

Velocity is measured in distance/second and acceleration is measured in distance/second².

5. From the given v – t graph figure, it can be inferred that the object is :

(a) In uniform motion
(b) At rest
(c) In non-uniform motion
(d) Moving with uniform acceleration

Solution

(a) In uniform motion

From the given graph it is clear that velocity of the object remains constant throughout the 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^–1. It implies that the boy is :
(a) At rest
(b) Moving with no acceleration
(c) In accelerated motion
(d) Moving with uniform velocity

Solution

(c) In accelerated motion

Boy is moving in a circular motion and circular motion is an 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^–1

Solution

(b) m

Area given in the graph represents displacement and its unit is meter.

8. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in figure.

Choose the correct statement :
(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.

Solution

(b) Car B is the slowest.

Graph shows that Car B covers the least distance in a given time.

9. Which of the following figure represents uniform motion of a moving object correctly?

Solution

(a)

Distance in graph (a) is uniformly increasing with time.

10. Slope of a velocity-time graph gives :
(a) The distance
(b) The displacement
(c) The acceleration
(d) The speed

Solution

(c) The 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.

Solution

(a) If the car is moving on straight road.

In all the other cases, displacement can be less than distance.

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 your answer.

Solution

When displacement is zero, it does not mean that the distance is zero. Distance can be zero when the object moves back to the place it started from. Displacement is either equal or less than distance. Distance is always greater than one and it cannot have a negative value.

13. How will the equations of motion for an object moving with a uniform velocity change?

Solution

In uniform motion, if acceleration (a) = 0
So in this case, 1st equation of motion becomes,
v = u + at
If a = 0 then v = u
If a = 0, 2^nd equation of motion becomes,
s = ut + ½at²
⇒ s = ut
If a = 0 , 3^rd equation of motion becomes,
v² – u² = 2as
⇒ v² – u² = 0

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 figure 8.4. Plot a velocity-time graph for the same.

Solution

From the graph, we get.
Initial velocity (u) = 0 (As time and displacement are zero.)
Velocity after 50 s, v = Displacement/Time
By substituting the values, we get
v = 100/50 = 2 ms^-1
Velocity after 100 s, v = Displacement/Time
Displacement = 0; Time = 100 s
Therefore,
By substituting the values, we get
v = 0/100 = 0

V	0	2	0
T	0	50	100

Velocity – time graph for above data:

15. A car starts from rest and moves along the x-axis with constant acceleration 5 ms–2 for 8 seconds. If it then continues with constant velocity, what distances will the car cover in 12 seconds since it started from the rest?

Solution

As the car moves from rest, therefore, u = 0, a = 5 ms2, t = 8 sec, distance (s) = ?
s’ = ut + ½at²
= 0 + ½ × 5 × (8)²
= 0 + 320/2
= 160 m
To calculate the velocity after 8 sec,
v = u + at
= 0 + 5 × 8 = 40 m/s
So, the distance travelled with this velocity for remaining 4 sec (12 s – 8s = 4s)
s” = 40 × 4 = 160 m
Therefore total distance travelled by the car
= s’ + s”
= 160 +160 = 320 m

16. A motorcyclist drives from A to B with a uniform speed of 30 km h^-1 and returns back with a speed of 20 km h^–1. Find its average speed.

Solution

Let AB = x,

17. The velocity-time graph figure shows the motion of a cyclist. Find
(i) its acceleration
(ii) its velocity and
(iii) the distance covered by the cyclist in 15 seconds.

Solution

(i) Since velocity is not changing, acceleration is equal to zero.
(ii) Through graph, velocity = 20 m s^–1
(iii) Distance covered in 15’ seconds,
s = u × t
= 20 × 15 = 300 m

18. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.

Solution

The velocity versus time graph of a stone thrown upwards vertically is as given below:

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 of 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

Initial difference in height
h = (150 – 100) m = 50 m
Both the objects are dropped simultaneously from rest (zero velocity),
Distance travelled by first body in 2 s
d₁ = ut + ½at²
Initial speed = 0 (as the object was at rest), t = 2s
Therefore, by substituting the values we get
= 0 + ½ g(2)² = 2g
= 2 × 9.8 = 19.6 m
Where, g = 9.8
Distance travelled by another body in
2s = d₂ = 0 + ½g(2)²
= 2g = 19.6 m
After 2s, height at which the first body will be
h₁ = 150 – 19.6
After 2s, height at which the second body will be
h₂ = 100 – 19.6
Thus, after 2s, difference in height
= h₁ – h₂
= 150 – 19.6 – (100 – 19.6)
= 50 m
After 2s, the difference in height will be 50 m = Initial difference in height (h)
Thus, difference in height does not vary with time.

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

Let’s acceleration of the body be ‘a’.

Since acceleration is the same, we have v′ =0+ (10 × 7) = 70 m s^–1

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	3	4	4	6	4	2	0

Solution

Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.

Average velocity for first 4s :
Average velocity = Change in displacement/Total time taken

22. An electron moving with a velocity of 5 × 10⁴ m s^–1 enters into a uniform electric field and acquires a uniform acceleration of 10⁴ m s^–2 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,
Initial velocity, u = 5 × 10⁴ ms^–1
and acceleration, a = 10⁴ m s^–2
(i) Final velocity = v = 2 u
v = 2 × (5 × 10⁴ ms^-1)
= 10 × 10⁴ ms^-1 ; t = ?
To find t, use v = u + at

23. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds. Using the equation of motion s = ut + ½ at² .

Solution

Using the equation of motion,

S = ut + ½at²

The distance travelled in 5 s,

Similarly, the distance travelled in 4 s,

Therefore, the distance travelled in the interval between 4thand 5thsecond = (S –S’)

24. Two stones are thrown vertically upwards simultaneously with their initial velocities u₁ and u₂ respectively. Prove that the heights reached by them would be in the ratio of (u₁)² : (u₂)² (Assume upward acceleration is – g and downward acceleration to be + g).

Solution 

For upward motion, v² = u² – 2gh