NCERT Exemplar Class 9 Science Chapter 8 Motion are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 8 Motion.

NCERT Exemplar Class 9 Science Solutions Chapter 8 Motion

Multiple Choice  Questions

Question 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) Displacement is the shortest distance measured from the initial position to final position.

Therefore, displacement after jralf a circle is the diameter of the circular path, i.e., 2r

Question 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) At highest point, v = 0
Using v²– u² = 2as, we get
0- u²=2(-g)h( ∵ a=-g and s=h)
or h=u²/2g

Question 3.
Thenumerical 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) As displacement is the shortest distance between two positions so it is generally less than distance. But displacement can be equal to distance when path taken is a straight line. So, the numerical ratio of displacement to distance for a moving object is equal or less than 1.

Question 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) As s α t², ∴ v α t and a = constant i.e., the object moves with uniform acceleration.

Question 5.
From the given v – t graph 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) As the given v-t graph shows that velocity is same for all time values, so, the object is in uniform motion.

Question 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
Solution:
(c) The merry-go-round moves in a circular path with constant speed, but its velocity changes due to the change in the direction of motion at every point. It implies that the boy on merry-go-round is in accelerated motion.

Question 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) Area under v-t graph represents distance and unit of distance is metre (m).

Question 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 6 is the slowest.
(c) Car O is faster than car C.
(d) Car C is the slowest.
Solution:

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

Solution:
(a) In uniform motion, object covers equal distances in equal intervals of time. Therefore, the distance-time graph must be a straight line inclined to time axis. Hence, graph (a) is correct.

Question 10.
Slope of a velocity – time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed
Solution:
(c) Slope of a velocity-time graph gives the rate of change of velocity or the acceleration.

Question 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, the distance moved and the magnitude of displacement are equal.

Short Answer Type Questions

Question 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:
In a given interval of time, displacement of a moving object is zero when its final position is the same as initial position, whereas distance travelled by the object is not zero. For example, an athlete moving on a circular track. If he starts from A and completes one round and reaches back to point A, his displacement is zero whereas distance travelled by him is not zero but 2πr.

Question 13.
How will the equations of motion for an object moving with a uniform velocity change?
Solution:
For an object moving with a uniform velocity, acceleration a = 0.

Question 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. Plot a velocity-time graph for the same

Solution:

Symmetry of graph shows that the girl comes back to her initial position with same velocity (2 m s^-1) but in opposite direction. So, the velocity-time graph will look like

Question 15.
A car starts from rest and moves along the x-axis with constant acceleration 5 m s~2 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:

Question 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:

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

Solution:
Velocity-time graph is a straight line parallel to time axis, so, velocity of the cyclist is constant.
(i) Acceleration = 0
(ii) At, f = 15 s velocity = 20 m s’1 (from the given graph)
(iii) Distance covered by the cyclist in 15 s
= Area under v-t graph during that time interval
= 20 m s’1 x 15 s = 300 m.

Question 18.
Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
Solution:
When a stone is thrown vertically upwards, its velocity at highest point is zero. As acceleration due to gravity (g) acts vertically downwards, so the upward motion of stone is uniformly decelerated and the downward motion is uniformly accelerated. This makes the velocity of stone while reaching at ground equal to initial velocity of stone. So the velocity-time graph will look like.

Here, PQ corresponds to upward motion and QR corresponds to downward motion of stone.

Long Answer Type Questions

Question 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:
Initially, difference in heights of two objects
= 150 m – 100 m = 50 m
Distance travelled by first object in 2 s

Difference in heights does oot vary with time as long as both the objects are in motion. However, when second object reaches ground and first one is still in motion, then it decreases.

Question 20.
An object starting from rest travels 20m in first 2 s and 160m in next 4 s. What will be the velocity after 7 s from the start.
Solution:
For first 2 s motion of object,
u = 0, f = 2 s, s = 20 m.

Question 21.
Using following data, draw time – displacement graph for a moving object:

Time (s)	P	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.
Solution:
From the given data, displacement-time graph is shown as

Question 22.
An electron moving with a velocity of 5 x 104 ms-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:

Question 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:
Let the object be moving with initial velocity u m s”1 and uniform acceleration a ms-²

Question 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 u2: uf (Assume upward acceleration is -g and downward acceleration to be +g
Solution:
At the highest point, v=0
For the sone thrown with velocity u₁