NCERT Exemplar Solutions Class 9 Science Motion

Question 3:  The numerical ratio of displacement and 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 3:   (d) equal or less than 1.

Since the degree of displacement is not equal to distance, the displacement of an object can be less than or equal to the distance covered by the object. It is possible, however, if the motion is in a straight line with no change in direction.

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 4:   (b) uniform acceleration.     

From second equation of motion, s = ut + ½ at²

If object start from rest i.e., initial velocity (u) = 0 and acnice an acceleration (a) in time (t)

Then,                               s = 0 × t + ½ at²

                                         s = ½ at²

                                          s  t², if a = constant

So, the object moves with constant or uniform acceleration.

Question 5:  From the given v-t graph (see 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 5:   (a) in uniform motion. 

The v-f graph clearly indicates that the object's velocity does not shift over time, suggesting that the object is travelling in a uniform manner.

Question 6:  Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms^{– a}. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity

Solution 6:   (c) in accelerated motion.

The speed of a merry-go-round stays constant, but the velocity varies as the direction changes, meaning that the motion is accelerating. As a result, we can assume that the boy is moving at a faster pace.

Question 7:  Area under v-t graph represents a physical quantity which has the unit
(a) m²                

(b) m                       

(c) m⁵                       

(d) ms^-1

Solution 7:   (b) m.

Area under v-t graph represent displacement whose unit is metre or (m).
Because, unit of velocity v = m/s and unit of time (T) = s.

Unit of (v – t) graph =  ^m/_s X s = m.

Hence, the unit of (v-t) graph is metre (m).

Question 8:  Four cars A, B, C and D are moving on a leveled road.
Their distances 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 8:   (b) Car B is the slowest. 

The speed is represented by the slope of the distance-time graph. It is obvious from the graph that the slope of the distance-time graph for car B is lower than for all other vehicles. As a result, the slope for car B is the smallest. As a result, car B is the slowest.

Solution 9: (a).
A straight line represents uniform motion on a distance-time graph (because in uniform motion object covers equal distance in equal interval of time).

Question 10:  Slope of a velocity-time graph gives

(a) the distance                                        

(b) the displacement

(c) the acceleration                                  

(d) the speed

Solution 10:  (c) the acceleration.

Slope of velocity-time graph gives acceleration.

Because slope of the curve =  V/t , where =V/t 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 11:   (a) If the car is moving on straight road. 

Only when moving along a straight line does the distance travelled and the degree of displacement equalize. Since the shortest path between the original and the find path is displacement.

As a result, the distance travelled and the degree of displacement is equal for a car travelling on a straight path.

Short Answer Type Questions........................

Question 12:  The displacement of a moving object in a given interval of a time is zero. Would the distance travelled by the object also be zero? Justify your answer?

Solution 12:   A moving object's displacement in a given interval is zero, i.e., the object returns to its original location in the given time (displacement is the shortest distance between the initial and final position of an object).

Since distance is the total length of the path travelled by the body, it will not be zero in this case. The length of the path travelled is not zero if the object returns to its original location.

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

Solution 13: 

We know that, the equations of uniformly accelerated motion are

(i) v = u + at               

(ii) s = ut +  at²

(iii) v² = u² + 2as

       Where,   u = Initial velocity

                       v = Final velocity

                       a = Acceleration

                       t = Time

         and       s = Distance
For an object moving with uniform velocity (velocity which is not changing with time), then acceleration
a = 0.
So, equations of motion will become (putting a = 0 in above equations)

i)        v =u                                       

ii)       s =ut

iii)     v² =u²

Question 14:  A girl walks along a straight path to drop a letter in the letter box and comes back to her initial position. Her displacement-time graph is shown in figure. Plot a velocity-time graph for the same.

Solution 14:

From the graph,

(i) Initial velocity,           u = 0         [Since, displacement and time is zero]

Question 15:  A car starts from rest and moves along the X-axis with constant acceleration 5 ms^-2 for 8 s. If it then continues with constant velocity. What distance will the car cover in 12 s, since it started from rest?

Solution 15:

Given, the car starts from rest, so its initial velocity u = 0
Acceleration, (a) = 5 ms^-2 and time (t) = 8 s
From first equation of motion,
                                             v = u + at
On putting a = 5 ms^-2 and t = 8 s in above equation, we get
                                        v = 0+ 5×8= 40ms’¹
So, final velocity v is 40 ms’¹.
Again, from second equation of motion,

                                        s = ut + ½ at²

on putting t = 8 s and a= 5 ms^-2 in above equation, we get

                           s = 0 × 8 + ½ × (8)² = ½ × 5 × 64 = 5 × 32 = 160 m     

So, the distance covered in 8s is 160 m.                                                                       
Given,           total time t = 12 s.
After 8s, the car continues with constant velocity i.e., the car will move with a velocity of 40 ms^-1.
So, remaining time t’= 12 s-8 s= 4s
The distance covered in the last 4s (s’) =Velocity x Time [∴ Distance = Velocity x Time]
                                                              = 40 x 4 = 160 m
[We have used the direct formula because after 8 s, car is moving with constant velocity i.e., zero acceleration].
Total distance travelled in 12 s from the start
                        D = s + s’= 160+ 160= 320 m.

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

Solution 16:

Let the distance between A and B be x km.

Time taken from A to B

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

Solution 17:

i)        Form the graph, it is clear that velocity is not changing with time i.e., acceleration is zero. 

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

Solution 18: A stone has some initial velocity as it is hurled vertically upwards (letu). The velocity of the stone decreases as it travels (v it is going towards gravity) until it reaches its highest point (maximum height). Enable the stone to travel to the highest point in time‘t’ seconds.
After that, the stone begins to fall (with zero initial velocity) and its velocity gradually increases (due to gravity) until it reaches its initial point of projection with velocity v in the same amount of time (with which it was thrown). So,

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 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 19:

Height of second object from the ground after 2s then h₂ = 100m – 20m = 80m Now, difference in the height after 2s = h₁ – h₂ = 130 – 80= 50 m

The difference in heights of the objects will remain same with time as both the objects have been dropped from rest and are falling with same acceleration i.e., (acceleration due to gravity).

Question 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 7s from the start?

Solution 20:

Given, object starts from rest, u = 0, t = 2s and s = 20m