NCERT Solutions for Class 9 Science Chapter 8 - Motion
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NCERT Solutions for Class 9 Science Chapter 8 Motion are provided here with simple step-by-step explanations. These solutions for Motion are extremely popular among class 9 students for Science Motion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 9 Science Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 9 Science are prepared by experts and are 100% accurate.

Page No 57:

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

Answer:

After half-circle particle will be diametrically opposite to its initial point. Therefore, the displacement of the particle will be 2r.

Hence, the correct answer is option C.

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Question 2:

A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) $\frac{u}{g}$
(b) $\frac{u^{2}}{2 g}$
(c) $\frac{u^{2}}{g}$
(d) $\frac{u}{2 g}$

Answer:

A body thrown upward with a velocity of u attains the greatest height when its final velocity becomes zero.

$2 a s = v^{2} - u^{2} 2 (- g) h = 0^{2} - u^{2} - 2 g h = - u^{2} 2 g h = u^{2} h = \frac{u^{2}}{2 g}$

Hence, the correct answer is option B.

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Question 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:

Since displacement is always less than or equal to the distance, the numerical ratio is equal to or less than 1.
Hence, the correct answer is option D.

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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

Answer:

$x α t^{2} x = k t^{2} \frac{x}{t^{2}} = k$

The unit of acceleration is metre per second square and the ratio of x and $t^{2}$ is constant. Therefore, the body moves with uniform acceleration.
Hence, the correct answer is option B.

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Question 5:

From the given v – t graph (Fig. 8.1), 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

Answer:

Since the velocity is constant, therefore the body is in uniform motion.
Hence, the correct answer is option A.

Page No 58:

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^–1. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity

Answer:

Since the boy is moving in circular path, its direction is changing at every point. Therefore his velocity is not uniform and he is in accelerated motion.
Hence, the correct answer is option C.

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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

Answer:

Area under velocity - time graph denotes displacement whose unit is m.
Hence, the correct answer is option B.

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Question 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

(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:

Slope of distance time graph denotes speed. Higher the slope the higher is the speed. Therefore, Car B is the slowest.
Hence, the correct answer is option B.

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Question 9:

Which of the following figures (Fig. 8.3) represents uniform motion of a moving object correctly?

Answer:

Slope of distance time graph denotes speed. For a body to have uniform motion its slope should be constant and should not be zero otherwise it will remain at rest.
Hence, the correct answer is option A.

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Question 10:

Slope of a velocity – time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed

Answer:

Slope of a velocity - time graph gives the acceleration.
Hence, the correct answer is option C.

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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

Answer:

The distance moved and the magnitude of displacement are equal if the car is moving on a straight road.
Hence, the correct answer is option A.

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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.

Answer:

It is not mandatory that if displacement of a moving object is zero then distance will also be zero. This is possible when a body revolves in a circular path and returns back to its initial position. In this scenario the displacement is zero but the body has covered some distance.

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Question 13:

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

Answer:

If a body is moving with uniform velocity then
$v = u$ and $a = 0$
Equations of motion will be
First equation: $v = u$
Second equation: $s = u t$
Third equation: $v^{2} - u^{2} = 0$

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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 Fig.8.4. Plot a velocity–time graph for the same.

Answer:

Velocity - time graph for the given motion is as follows

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Question 15:

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

Answer:

Initial velocity ( $u$ ) = 0
Acceleration ( $a$ ) = 5 $m / s^{2}$
Time ( $t$ ) = 8 s

Distance covered in first 8 seconds
$s = u t + \frac{1}{2} a t^{2} s = (0 \times 8) + \frac{1}{2} \times 5 \times 8^{2} s = 160 m$

Velocity attained by body after 8 seconds
$v = u + a t v = 0 + (5 \times 8) v = 40 m / s$

Distance covered by body in next 4 seconds
$s = v t s = 40 \times 4 s = 160 m$

Total distance covered by body in first 12 seconds
$s = 160 m + 160 m s = 320 m$

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Question 16:

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

Answer:

Let the distance between the two places A and B be $x$ km
Time taken by motorcyclist to go from A to B is
$t_{1} = \frac{x}{30} h$

Time taken by motorcyclist to return from B to A is
$t_{2} = \frac{x}{20} h$

Total distance covered by motorcyclist is $2 x$
Total time taken by him is
$T = \frac{x}{30} + \frac{x}{20}$

Average speed (v) is

$v = \frac{Total distance covered}{Total time taken} v = \frac{2 x}{(\frac{x}{30} + \frac{x}{20})} v = \frac{2 x}{(\frac{2 x + 3 x}{60})} v = \frac{2 x}{(\frac{5 x}{60})} v = 24 km h^{- 1}$

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Question 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.

Answer:

(i) Since velocity time graph is parallel to time axis, velocity is constant hence acceleration is zero.

(ii) From velocity - time graph velocity is constant which is equal to 20 m/s.

(iii) Distance covered by cyclist is given by the area under velocity - time graph.

Distance ( $s$ ) is

$s = v \times t s = 20 \times 15 s = 300 m$

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Question 18:

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

Answer:

Considering upward direction as positive and downward direction as negative, velocity - time graph is as follows

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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?

Answer:

Initially, difference in the height of two objects is 50 m
Distance covered by the first object in 2 s
$s_{1} = u t + \frac{1}{2} a t^{2} s_{1} = (0 \times 2) + (\frac{1}{2} \times 10 \times 2^{2}) s_{1} = 20 m$

Height of the first object from the ground after 2 s
$h_{1} = 150 m - 20 m h_{1} = 130 m$

Similarly, distance covered by second object in 2 s
$s_{2} = u t + \frac{1}{2} a t^{2} s_{2} = (0 \times 2) + (\frac{1}{2} \times 10 \times 2^{2}) s_{2} = 20 m$

Height of the second object from the ground after 2 s
$h_{2} = 100 m - 20 m h_{2} = 80 m$

Difference in the height of two objects after 2 s
$h = 130 m - 80 m h = 50 m$

Difference in the height of two objects will remain same as long as both the objects are in air. But once second object reaches the ground then distance between the two objects will keep on decreasing and eventually will become zero.

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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 7 s from the start.

Answer:

Time ( $t$ ) = 2 s
distance ( $s$ ) = 20 m
Initial velocity ( $u$ ) = 0 m/s
Applying second equation of motion:
$s = u t + \frac{1}{2} a t^{2}$
$s = (0 \times 2) + (\frac{1}{2} \times a \times 2^{2}) 20 = 2 a a = 10 m / s^{2}$

Velocity of the body after 7 s from the start
$v = u + a t v = 0 + (10 \times 7) v = 70 m / s$

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Question 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:

Displacement time graph for the following data is as shown

Average velocity for the first 4 seconds

$v = \frac{4 - 0}{4 - 0} v = 1 {ms}^{- 1}$

Average velocity for the next 4 seconds

$v = \frac{4 - 4}{8 - 4} v = 0 {ms}^{- 1}$

Average velocity for the last 6 seconds

$v = \frac{0 - 6}{16 - 10} v = \frac{- 6}{6} v = - 1 {ms}^{- 1}$

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Question 22:

An electron moving with a velocity of 5 × 10⁴m s^–1 enters into a uniform electric field and acquires a uniform acceleration of 10⁴m s^–2in 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?

Answer:

(i) Initial velocity ( $u$ ) = $5 \times 10^{4} {ms}^{- 1}$
Acceleration ( $a$ ) = $10^{4} {ms}^{- 2}$
Final velocity ( $v$ ) = $2 u$

Applying first equation of motion:
$v = u + a t$

2 u = u + a t a t = u t = \frac{u}{a} t = \frac{5 \times 10^{4}}{10^{4}} t = 5 s

(ii) Distance (s) covered in 5 s is

2 a s = v^{2} - u^{2} 2 \times (10^{4}) s = {(2 u)}^{2} - u^{2} 2 \times (10^{4}) s = 4 u^{2} - u^{2} 2 \times (10^{4}) s = 3 u^{2} s = \frac{3 {(5 \times 10^{4})}^{2}}{2 (10^{4})} s = 3.75 \times 10^{5} m

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Question 23:

Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4^thand 5^thseconds.

Answer:

Let the initial velocity of the body be $u$ and acceleration of the body be $a$ .

Distance covered by an object in 4 s
$s_{1} = u t + \frac{1}{2} a t^{2} s_{1} = 4 u + \frac{1}{2} \times a \times 4^{2} [∴ t = 4 s] s_{1} = 4 u + 8 a$

Distance covered by an object in 5 s

$s_{2} = u t + \frac{1}{2} a t^{2} s_{2} = 5 u + \frac{1}{2} \times a \times 5^{2} s_{2} = 5 u + \frac{25 a}{2}$

Distance covered by an object between $4^{t h}$ and $5^{t h}$ second is

$s = s_{2} - s_{1} s = (5 u + \frac{25 a}{2}) - (4 u + 8 a) s = (u + \frac{9 a}{2}) m$

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Question 24:

Two stones are thrown vertically upwards simultaneously with their initial velocities u₁and u₂respectively. Prove that the heights reached by them would be in the ratio of $u_{1}^{2} : u_{2}^{2}$ (Assume upward acceleration is –g and downward acceleration to be +g).

Answer:

Let the height attained by first stone be h₁:
Using third equation of motion, $2 a s = v^{2} - u^{2}$
$\Rightarrow 2 (- g) h_{1} = 0^{2} - {u_{1}}^{2} \Rightarrow - 2 g h_{1} = - {u_{1}}^{2} \Rightarrow h_{1} = \frac{{u_{1}}^{2}}{2 g}$

Similarly, height attained by second stone is h₂.
$\Rightarrow 2 (- g) h_{2} = 0^{2} - {u_{2}}^{2} \Rightarrow - 2 g h_{2} = - {u_{2}}^{2} \Rightarrow 2 g h_{2} = {u_{2}}^{2} \Rightarrow h_{2} = \frac{{u_{2}}^{2}}{2 g}$

Ratio of heights attained by two stones is

$\Rightarrow \frac{h_{1}}{h_{2}} = \frac{(\frac{{u_{1}}^{2}}{2 g})}{(\frac{{u_{2}}^{2}}{2 g})} \Rightarrow \frac{h_{1}}{h_{2}} = \frac{{u_{1}}^{2}}{{u_{2}}^{2}}$

Hence, the ratio of heights reached by them is in the ratio of $u_{1}^{2} : u_{2}^{2}$ .

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