# NCERT Solutions Class 9 Science Chapter 8 Motion

## NCERT Solutions Class 9 Science Chapter 8 Motion

NCERT Solutions Class 9 Science Chapter 8 Motion: National Council of Educational Research and Training Class 9 Science Chapter 8 Solutions – Motion. NCERT Solutions Class 9 Science Chapter 8 PDF Download.

### NCERT Solutions Class 9 Science Chapter 8: Overview

 Board NCERT Class 9 Subject Science Chapter 8 Chapter Name Motion Topic Exercise Solutions

### NCERT Solutions Class 9 Science Chapter 8 – Motion

Part 1

1.) An object has moved through adistance. Can it have zero displacement?If yes, support your answer with an example.

Answer: Yes, displacement can also zero. If the object moves some distance and returned to it’s original position then displacement becomes zero. The value of displacement in cyclic motion also zero.

Example- The car moves 40 km in a straight line and returned back to it’s initial position then the value of distance is 80 km and the value of displacement becomes zero.

2.) A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer: Given, side of square= 10 m,

Perimeter= 40 m,

The farmer covers 40 m in 40 seconds.

Thus he covers 10 m in 10 seconds.

Time  = 2 minutes 20 seconds = 140 seconds.

Now we have to calculate displacement covered in 140 seconds. After 140 seconds he complete three rounds and after that he reach at point T as shown in figure.

We know that, displacement is the shortest distance between initial and final position. Initially he was at point R and finally he is at point T.

To calculate displacement,  we use Pythagoras theorem,

(RT)² =( RS)²+ (ST)²

(RT)² = 10² + 10²

(RT)² = 100 + 100

(RT)² = 200

(RT)² = 2×100

Taking square root on both side,

RT = 10(2)1/2

So the magnitude of displacement is  10(2)1/2m.

3.) Which of the following is true for displacement?

(a) It cannot be zero.

(b) Its magnitude is greater thanthe distance travelledby theobject.

(a) It cannot be zero.

False, displacement can also zero. The value of displacement in cyclic motion is zero.

(b) Its magnitude is greater than the distance travelled by the object.

False, The magnitude of displacement is always less than or equal to the distance travelled by the object.

Part 2

1) Distinguish between speed and velocity.

Answer: The difference between seed and velocity is as,

• Speed is scalar quantity and velocity is vector quantity.
• Seed has only magnitude but velocity has both magnitude and direction.
• Speed can not zero or negative but velocity can zero or negative also.
• If the particle moves in a straight line without changing the direction then magnitude of velocity and speed are equal.
• Speed is depend on path length but velocity does not depend on path length.

2) Under what condition (s) is the magnitude of average velocity of an object equal to its average speed?

Answer: If an object moves in a straight line without changing it’s direction then the magnitude of average velocity of an object equal to its average speed.

Example: A bus moves 50 km in a straight road in 2 hrs.

In this case the value of distance and displacement is same and time also same thus average speed and average velocity also becomes same.

3) What does the odometer of an automobile measure?

Answer: Odometer of an automobile measures speed of that vehicle.

4) What does the path of an object look like when it is in uniform motion?

Answer: When the object perform uniform motion then it path is represented by straightline.

5) During an experiment, a signal from a spaceship reached theground station in five minutes. What was the distance of thespaceship from the groundstation? The signal travels at thespeed of light, that is, 3 × 10⁸ m/s.

Answer: Given, speed of signal = 3 × 10⁸ m/s, time to reach signal at ground = 5 minutes.

Time = 5 × 60              (1 minute = 60 seconds)

Time = 300 seconds.

As we know,

Speed = distance/time

Distance = speed × time

Distance = 3 ×10⁸ × 300

Distance = 9 × 10¹⁰ m.

So the spaceship covers 9 × 10¹⁰ m distance.

Part 3

1.) When will you say a body is in

(i) uniform acceleration?

(ii) non-uniform acceleration?

i) If the object moves with constant velocity then we say that the motion is uniform motion. It is zero accelerated motion.

Example: When a bus travel 40 km in 2 hrs and next 80 km in 4 hrs then we say the bus travels with uniform motion.

ii) If the object moves acceleration then we say that the motion is non uniform motion. It is accelerated motion.

Example- When a bus travel 40 km in 2 hrs and next 80 km in 3 hrs then we say the bus travels with non uniform motion.

2.) A bus decreases its speed from80 km/h to 60 km/h in 5 s. Find the acceleration of thebus.

Answer: Given, initial speed = u = 80 km/h,

u = 80 × 5 /18 m/s = 22.22 m/s.

final speed = v = 60 km/h,

v = 60 × 5 /18 m/s = 16.66 m/s.

Time = 5 seconds,

As we know the definition of acceleration,

Acceleration = (v – u)/t

Acceleration = (16.66-22.22 )/5

Acceleration = -5.56/5

Acceleration = -1.112 m/s²

The acceleration of the bus is -1.112m/s².

3.) A train starting from a railwaystation and moving with uniformacceleration attains a speed40 km/h in 10 minutes. Find itsacceleration.

Answer: Train starting from rest so initial speed = u = 0 m/s,

final speed = v = 40 km/h,

v = 40 × 5 /18 m/s = 11.11 m/s.

Time = 10 minutes = 600 seconds.

As we know the definition of acceleration,

Acceleration = (v – u)/t

Acceleration = (11.11-0)/600

Acceleration = 11.11/600

Acceleration = 0.0185 m/s²

The acceleration of the train is 0.0185 m/s².

Part 4

1.) What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer:  The nature of distance-time graph for uniform motion of object is straight line and the nature of distance-time graph for non-uniform motion of object is not straight line.

2.) What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer: If the motion of an object whose distance-time graph is a straight line parallel to the time axis then we say that the object is at rest. If the graph is parallel to time axis then the particle does not change it’s position.

3.) What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Answer: If the motion of an object whose speed-time graph is a straight line parallel to the time axis then the we say that the object does not change it’s speed and such motion is called as uniform motion.

4.) What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer: As we know that area is the product of two dimensions of the object.

Area under velocity-time graph= velocity × time.  (1)

But we know that

Velocity = displacement/time

Displacement = velocity × time.   (2)

From equation 1 and 2 we say that,

Area under velocity-time graph= displacement

The area occupied below the velocity-time graph is displacement.

Part 5

Part 5

1.) A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find

(a) the speed acquired,

(b) the distance travelled.

Answer: A bus starting from rest so initial speed = u = 0 m/s,

Acceleration = a = 0.1 m/s², time = t = 2 minutes,

t = 120 seconds.

(a) As we know,

Acceleration = (v-u)/t

0.1 = (v-0)/120

0.1 = v/120

0.1 × 120 = v

v = 12 m/s.

The speed acquired is 12 m/s.

(b) As we know that, Second equation of kinematics,

x = ut + ½ at²

x = 0(20) + ½( 0.1× 120²)

x = ½ (0.1 × 14400)

x = 0.1 × 7200

x = 720 m.

The distance travelled by bus is 720 m.

2.) A train is travelling at a speedof 90 km/h. Brakes are appliedso as to produce a uniform acceleration of – 0.5 m/s². Findhow far the train will go before itis brought to rest.

Answer: Given, initial speed= u = 90 km/h = 90× 5/18 m/s,

Initial speed= 25 m/s,

Acceleration = -0.5 m/s²,

finally it stop so final speed = v = 0 m/s.

As we know that third equation of motion,

v² = u² + 2ax

0² = 25² + 2(-0.5)x

x = 25²

x = 625 m.

The train covers 625 m before stop.

3.) A trolley, while going down aninclined plane, has anacceleration of 2 cm/s. What willbe its velocity 3 s after the start?

Answer: Given, acceleration= a = 2 cm/s²,

But 1 m = 100 cm, so

Acceleration = 2cm/s² = 0.02 m/s², time = t= 3 s,

Initial it stop from rest so initial velocity= v = m/s,

We calculate final velocity using 1st equation of motion,

v = u +at

v = 0 + 0.02 × 3

v = 0.06 m/s

The velocity of trolley after 3 seconds will be 0.06 m/s.

4.) A racing car has a uniformacceleration of 4 m/s². What distance will it cover in 10 s after start?

Answer: Given, acceleration= a = 4 m/s², time =t = 10 s,initial speed of the car is zero because initially it was at rest.

According to second equation of motion,

s = ut + ½ at²

s = 0(10) + ½ (4) ×10²

s = 2 ×100

s = 200 m.

The racing car covers 200 m in 10 seconds.

5.) A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If theacceleration of thestone during its motionis 10 m/s² in the downward direction, whatwill be the height attained by thestone andhow much time will ittake to reach there?

Answer: Given, initial speed of stone = 5 m/s,

Finally it stop when reach at maximum height, v = 0 m/s,

Acceleration opposes the motion, a = -10 m/s²,

Firstly we calculate height using 3rd equation of

v² = u² +2as

0² = 5² + 2 (-10)s

20s = 25

s = 25/20

s = 1.25 m.

The stone attain height of 1.25 m.

Now we calculate time required to attain this height using 1st equation of motion,

v = u + at

0 = 5 +(-10)t

10t = 5

t = 5/10

t = 0.5 second.

It reach at 1.25 m in 0.5 s.

Motion exercise

1.) An athlete completes one round of a circular track of diameter200 m in 40 s. What will be the distance covered and thedisplacement at the end of 2 minutes 20 s?

Answer: Given, diameter = 2R = 200 m,

R = 100 m.

The athlete completes one round of circular track in 40 s. Therefore that athlete complete 3 and ½ round in 2 minutes 20 s.

Total distance covered in one round = 2 πR.

Total distance covered in one round = 2 × 3.142 ×100

Total distance covered in one round = 628.4 m.

The athlete completes 3 and ½ round in 2 minutes 20 seconds. So total distance covered in 3 ½  round = 3.5 × 628.4

Total distance covered in 3 ½ round = 2199.4 m.

Now we calculate displacement. The athlete reach at point B after 2 minutes 20 s. The displacement of the athlete is the diameter of that circular track.

Hence value of displacement after 2 minutes 20 seconds is 200 m

2.) Joseph jogs from one end A to the other end B of astraight300 m road in 2 minutes 30 seconds and thenturns aroundand jogs 100 m back to point C in another 1minute. What areJoseph’s average speeds and velocitiesin jogging (a) from A toB and (b) from A to C? a) The distance from A to B = 300 m, time to travelled 300 m = 2 minutes 30 seconds.

Time = 150 seconds.

Average speed = distance from A to B/time.

Average speed = 300/150

Average speed = 2 m/s.

The particle does not change the direction so value of speed and velocity are equal.

Thus the value of average velocity also 2 m/s.

b) Now we calculate average speed from A to C.

Total distance = 300 + 100

Total distance = 400 m.

Total time = Time required for move from A to B + Time required to move from B to C.

Total time = 150+ 60

Total time= 210 seconds.

Average speed = total distance/total time

Average speed = 400/210

Average speed = 1.90 m/s.

The particle changes the direction so value of speed and velocity are unequal.

We can calculate the value of displacement from the above figure.

Thus the value of displacement from A to C is 200 m.

Average velocity = total displacement/time

Average velocity = 200/ 210

Average velocity = 0.952 m/s.

3.) Abdul, while driving to school, computes the average speed forhis trip to be 20 km/h.  On his return trip along the sameroute, there is less traffic and the average speed is30 km/h.  What is the averagespeed for Abdul’s trip?

Answer: Given, average speed for forward journey = v1 = 20 km/h, average  speed for returned journey = v2 = 30 km/h.

If the particle travels same distance with different speed then the formula for calculating average speed in entire journey is,

Average speed = (2 v1 v2) /(v1+v2 )

Average speed = (2 ×20×30) /( 20+30)

Average speed = 1200 /50

Average speed = 24 Km/h.

The average speed of Abdul’s journey is 24 Km/h.

4.) A motorboat starting from rest on a lake accelerates in a straightline at a constant rate of 3.0 m/s² for 8.0 s. How far does theboat travel during this time?

Answer: Given, the motorboat starting from rest so initial speed = u = 0 m/s,

Acceleration = a = 3 m/s², time = t = 8 seconds.

The second equation of motion is as,

x = ut + ½ at²

x = 0× 3 + ½ × 3 × 8²

x = 0 + ½ 3× 64

x = 96 m.

The boat travels 96 m in 8 seconds.

5.) A driver of a car travelling at 52 km/h applies the brakes andaccelerates uniformly in the opposite direction. The car stopsin 5 s. Another driver going at 3 km/h  in another car applieshis brakes slowly and stops in 10 s. On the same graph paper,plot the speed versus time graphs for the two cars. Which ofthe two cars travelled farther after the brakes were applied?

Answer: The initial speed of car A = 52 km/h and stops after 5 seconds.

The initial speed of car B is 3 km/h and it stop after 10 s. We plot a graph for these two are as, We calculate distance travelled by object A after break.

Initial peed of A = 52 km/h = 52 × 5/18 m/s

Initial peed of A = 14.4 m/s.

Time = 5 s

As we know that area of speed time graph is the distance covered by the object.

Distance = area enclose by object A

But area enclosed by A is a triangle

Distance = ½ (base × height)

Distance = ½ (5 × 14.4)

Distance = 2.5 ×14.4

Distance = 36 m.

The distance travelled by A is 36 m.

Now we calculate distance travelled by B.

Initial peed of B = 3 km/h = 3 × 5/18 m/s

Initial peed of B = 0.833 m/s.

Time = 10 s

As we know that area of speed time graph is the distance covered by the object.

Distance = area enclose by object B

But area enclosed by B is a triangle

Distance = ½ (base × height)

Distance = ½ (10 × 0.833)

Distance = 5 × 0.833

Distance = 4.16 m.

The distance travelled by B is 4.16 m.

Car A travelled much distance after break than B.

6.) Fig 8.11 shows the distance-time graph of three objects A,Band C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?

(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?

(d) How far has B travelled by the time it passes C?

Answer: The above graph is between time and distance.

(a) The slope of distance time graph is the speed .

The slope object B is greater than other two objects so speed of B if greater than object A and C. Thus the object B is travelling fastest.

(b) Object B and C meet at 5.5 km. Object B and object A meet at 9.5 km. Object A and object C meet at 10.5 Km. These three objects do not meet at the same time on the road.

(c) Object C reach at 7.5 km when B passes A.

(d) B is at 4.5 km by the time it passes C.

7.) A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocitywill it strike the ground? After what time will it strike theground?

Answer: Given height = 20 m, the ball is dropped so initial velocity = 0 m/s,

Acceleration = a = 10 m/s²10

As we know the 3rd equation of kinematics,

v² = u² + 2 ah

v² = 0² + 2 × 10 × 20

v² = 400

Taking square root on both side,

v = 20 m/s.

Now we calculate time required the ball to reach the ground using 1st equation of motion,

v = u + at

20 = 0 + 10× t

t = 20/10

t = 2 seconds.

The final velocity of that stone is 20 m/s and time required to reach it to ground is 2 seconds.

8.) The speed-time graph for a car is shown is Fig. 8.12.

(a) Find how far does the car travel in the first 4 seconds. Shade thearea on the graph that represents the distancetravelled by the carduring the period.

(b) Which part of the graph represents uniform motion ofthe car?

a)

• To find the total distance covered by the car in first 4 seconds, we have to find the area under curve i.e. area of the shaded portion of graph.
• Initially we find area of each rectangle included in that portion.
• There are 5 divisions in each 2 seconds on X-axis.

So length will be 2/5 unit.

• And there are 3 divisions in each of 2m/s on Y-axis.

• Thus, area of each rectangle is given by,

Area= length × breadth= (2/5)×(2/3 ) = 4/15

• Now, we count the total number of rectangle, more than half rectangle, half rectangle and other will be neglected.
• Here, there are 57 full rectangle, 3 more than half rectangle, 3 half rectangle and other are negligible to count.

So, for 57 full rectangle

Area= 57×(4/15) = 15.2

• For 3 more than half rectangle,

Area= 3×(4/15) = 0.8

• And for 3 half rectangle,

Area= 3×(1/2)×(4/15) = 0.4

• Thus, area under the curve shaded= area of 57 full rectangle+ area of 3 more than half rectangle+ area of 3 half rectangle

= 15.2 + 0.8 + 0.4 = 16.4 square unit.

Thus, we can say that the distance covered by the car in first 4 seconds is 16.4 m.

b)

After time t= 6 seconds the speed is constant although time is increasing. So after t= 6 s, the graph shows uniform motion of car.

9.) State which of the following situations are possible and give an example for each of these:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving with an acceleration but with uniform speed.

(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) Yes, it is possible that the object with constant acceleration but with uniform speed. If we throw a stone in upward direction then it reach the maximum height. When the object is at maximum height then it has constant acceleration due gravity but zero velocity.

(b) Yes, it is possible that the object moving with an acceleration but with uniform speed. When the object perform uniform circular motion then it moves with constant speed and acceleration.

(c) Yes, it is possible that an object moving in a certain direction with an acceleration in the perpendicular direction. It is possible when an object perform circular motion.

10.) An artificial satellite is moving in a circular orbit of radius42250 km. Calculate its speed if it takes 24 hours to revolvearound the earth.

Answer: Given, R = 42250 km = 42250000 m, time  = 24 hours = 24 × 60 ×60

Time = 86400 seconds.

If a particle moves along circular track then we can calculate speed using formula,

Speed = 2 πR/t

Speed =( 2 × 3.142 × 42250000)/86400

Speed = 265499000/86400

Speed = 3072 m/s.

The speed of that satellite is 3072 m/s.

Updated: August 14, 2021 — 12:21 am