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# Motion - NCERT Solutions of Chapter 8 (Science) for Class 9 ## NCERT Solutions for Class 9 Science Chapter 8 - Motion

### Important Topics

Here are the important topics from the chapter which should should no way miss:

1. Motion

2. Describing Motion

3. Measuring the Rate of Motion

4. Rate of Change of Velocity

5. Graphical Representation of Motion

6. Equations of Motion by Graphical Method

7. Uniform Circular Motion

### Key Features of NCERT Solutions for Class 9 Science Chapter 8 - Motion

• NCERT Solutions uses a simple and easy-to-understand approach to teach students about various topics.

• All of the questions in the relevant NCERT textbooks are entirely solved in this book.

• To assist students in their preparations, NCERT Solutions provides complete answers to all of the questions.

• These answers will come in handy for Science Olympiads, CBSE Term I exams, and other competitive exams.

### Important Points

• A reference point or origin is required to explain the position of an object. To one observer, an object may appear to be moving while to another, it appears to be motionless.

• For instance, a passenger onboard a bus perceives the other passengers to be at rest, yet an observer outside the bus perceives the people to be moving.

• A convention, or a standard reference point or frame, is required to make observations easier. The reference frames of all objects must be the same.

The topic of motion in physics falls under the category of classical mechanics. Students are taught about this topic in ch 8 science class 9. But what if you didn’t get NCERT Solutions of chapter 8 science class 9 the first time that you read the chapter? If that is the case, then you are in the right place. In this article, students will learn everything that they need to know for writing NCERT solutions for class 9 science chapter 8.

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See More ## Access NCERT Solutions for Class 9 Science Chapter 8 – Motion

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

Ans: Yes. An object can have zero displacement if it has moved through a distance. Displacement is defined as the shortest distance from the initial point to the final point.

Hence, if the starting (initial) point is the same as the final point then the displacement of the object is zero.

Suppose a man is walking in a square park of length $20m$. He starts from point A and walks along all the corners of the park through points B, C and D and comes back to the same point A.

The total distance covered by the man$=20m+20m+20m+20m=80m$.

As the starting point and final point are same, the shortest distance between his initial and final position is zero

Therefore, the displacement is zero.

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

Ans: It is given that,

Farmer takes $40s$ to cover a square field of side $10m$.

$\Rightarrow Dis\tan ce=4\times 10=40m$

It is known that,$Speed=\frac{Dis\tan ce}{Time}$

$\Rightarrow Speed=\frac{40}{40}=1$

Therefore, speed of the farmer is $1m/s$.

In $2$minutes $20$seconds distance travelled is $Speed\times Time$.

$\Rightarrow Dis\tan ce=1\times (2\times 60+20)$

$\Rightarrow Dis\tan ce=140m$

Number of rounds farmer covered$=\frac{140}{40}=3.5$

After $2$minutes $20$seconds the farmer will be at the opposite end of starting point, completing $3$ and half rounds.

1. If the farmer starts from any corner of the field: The displacement will be equal to the diagonal of the field.

$\Rightarrow Displacement=\sqrt{{{10}^{2}}+{{10}^{2}}}=14.14m$

1. If the farmer starts from the middle point of any side of the field: The final point will be the middle point of the side opposite to the initial point.

$\Rightarrow Displacement=10m$

Therefore, the magnitude of displacement if the farmer starts at any corner is $14.14m$ and if the farmer starts from middle point of any side is $10m$.

3. Which of the following is true for displacement?

1. It cannot be zero.

Ans: Not true. When the initial and final position of the object is the same, then the displacement is zero.

1. Its magnitude is greater than the distance travelled by the object.

Ans: Not true. Displacement is the measure of the shortest distance between the initial and final position of an object.

Therefore, it is always smaller than or equal to the magnitude of distance travelled by the object.

4. An artificial satellite is moving in a circular orbit of radius $\mathbf{42250km}$. Calculate its speed if it takes $24$ hours to revolve around the earth?

Ans: It is given that,

Radius of the circular orbit, $r=\mathbf{42250km}$

Time taken by the satellite to revolve around earth, $t=24h$

Speed of the artificial satellite, $v=?$

It is known that,

$v=\frac{2\pi r}{t}$

$\Rightarrow v=\frac{2\times 3.14\times 42250}{24}$

$\Rightarrow v=1.105\times {{10}^{4}}km/h$

$\Rightarrow v=\frac{1.105\times {{10}^{4}}}{3600}km/s$

$\Rightarrow v=3.069km/s$

Therefore, the speed of the artificial satellite is $v=3.069km/s$.

5. Distinguish between speed and velocity.

Ans: The differences between speed and velocity are as follows:

 Speed Velocity The distance travelled by an object in a given interval of time is speed.Speed does not have any direction.Speed is either positive or zero but not negative. The displacement of an object in a given interval of time is velocity.Velocity has a unique direction.Velocity can be negative, positive or zero.

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

Ans: It is known that,

$Average\text{ }speed=\frac{Total\text{ }distance\text{ }covered}{Total\text{ }time\text{ }taken}$

$Average\text{ }velocity=\frac{Displacement}{Total\text{ }time\text{ }taken}$

Therefore, the magnitude of average velocity of an object is equal to its average speed when total distance covered is equal to the displacement.

7. What does the odometer of an automobile measure?

Ans: The distance covered by an automobile is recorded by the odometer of an automobile.

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

Ans: An object has a straight-line path when it is in uniform motion.

9. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is,$3\times {{10}^{8}}m{{s}^{-1}}$.

Ans: It is given that,

Time taken by a signal to reach ground from a spaceship $=5\min =5\times 60=300\sec$

Speed of the signal is equal to speed of light$=3\times {{10}^{8}}m{{s}^{-1}}$

It is known that,

$Speed=\frac{Distance\text{ travelled}}{Time\text{ }taken}$

$\Rightarrow Distance\text{ travelled = Speed}\times \text{Time taken}$

$\Rightarrow Distance\text{ travelled = 3}\times \text{1}{{\text{0}}^{8}}\times 300=9\times {{10}^{10}}m$

Therefore, the distance of the spaceship from the ground station is $9\times {{10}^{10}}m$.

10. When will you say a body is in

1. uniform acceleration?

Ans: When the magnitude and the direction of acceleration of a body is constant i.e., velocity changes at an equal rate then the body is said to be in uniform acceleration.

1. non-uniform acceleration?

Ans: When the acceleration of a body changes in magnitude or direction or both i.e., velocity changes at an unequal rate then the body is said to be in non-uniform acceleration.

11. A bus decreases its speed from $80km{{h}^{-1}}$to $60km{{h}^{-1}}$in $5s$. Find the acceleration of the bus.

Ans: It is given that,

Initial speed of the bus, $u=80km/h$

$\Rightarrow u=80\times \frac{5}{18}m/s=22.22m/s$

Final speed of the bus, $v=60km/h$

$\Rightarrow v=60\times \frac{5}{18}m/s=16.66m/s$

Time taken to decrease speed, $t=5s$

It is known that,

Acceleration,$a=\frac{v-u}{t}$

$\Rightarrow a=\frac{16.66-22.22}{5}$

$\Rightarrow a=-1.112m/{{s}^{2}}$

Therefore, the acceleration of the bus is $-1.112m/{{s}^{2}}$. The negative sign indicates that the velocity of the car is decreasing. Decreasing acceleration is called retardation.

12. A train starting from a railway station and moving with uniform acceleration attains a speed $40km/h$ in $10$ minutes. Find its acceleration.

Ans: It is given that,

Initial velocity of the train, $u=0$ (Train is starting from rest)

Final velocity of the train, $v=40km/h$

$\Rightarrow v=40\times \frac{5}{18}m/s=11.11m/s$

Time taken, $t=10\times 60=600s$

It is known that,

Acceleration,$a=\frac{v-u}{t}$

$\Rightarrow a=\frac{11.11-0}{600}$

$\Rightarrow a=0.0185m/{{s}^{2}}$

Therefore, the acceleration of the train is $0.0185m/{{s}^{2}}$.

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

Ans: The distance-time graph for uniform motion of an object is a straight line.

The distance-time graph for non-uniform motion of an object is a curved line.

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

Ans: A straight line parallel to the x-axis in a distance-time graph indicates that the position of the object does not change with time.

Therefore, the object is said to be at rest.

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

Ans: A straight line parallel to the time axis in a speed-time graph indicates that the speed of the object does not change with time.

Therefore, the object is moving uniformly.

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

Ans: The area of the velocity-time graph is displacement.

Consider the following figure which shows the velocity-time graph of a uniformly moving body.

Let, the velocity of the body at time $t$ be $v$.

Area of the shaded region$=Length\times Breadth$

Where,

$Length=l$

$Breadth=v$

$\Rightarrow Area=vt=velocity\times time$   ……$(1)$

It is known that,

$Velocity=\frac{Displacement}{time}$

$\Rightarrow Displacement=Velocity\times Time$  ……$(2)$

From equations $(1)$and $(2)$

$\Rightarrow Area=Displacement$

Hence, the area occupied below the velocity-time graph measures the displacement of the body.

17. A bus starting from rest moves with a uniform acceleration of $0.1m/{{s}^{2}}$for $2$ minutes. Find

1. The speed acquired

Ans: It is given that,

Initial velocity of the bus, $u=0$ (Bus is starting from rest)

Acceleration of bus, $a=0.1m/{{s}^{2}}$

Time taken, $t=2\min =120\sec$

Final velocity of the bus, $v=?$

It is known that,

Acceleration,$a=\frac{v-u}{t}$

$\Rightarrow 0.1=\frac{v-0}{120}$

$\Rightarrow v=12m/s$

Therefore, the speed acquired is $v=12m/s$.

1. The distance travelled

Ans: It is known that,

From, third equation of motion:${{v}^{2}}-{{u}^{2}}=2as$

$\Rightarrow {{(12)}^{2}}-{{(0)}^{2}}=2\times (0.1)\times s$

$\Rightarrow 144=0.2s$

$\Rightarrow s=720m$

Therefore, the distance travelled is $s=720m$.

18. A train is travelling at a speed of $90km{{h}^{-1}}$. Brakes are applied so as to produce a uniform acceleration of $-0.5m{{s}^{-2}}$. Find how far the train will go before it is brought to rest.

Ans: It is given that,

Initial speed of a train, $u=90km/h$

$\Rightarrow u=90\times \frac{5}{18}=25m/s$

Final speed of the train, $v=0$ (Train comes to rest finally)

Acceleration of train, $a=-0.5m/{{s}^{2}}$

Distance covered by the train, $s=?$

It is known that,

From, third equation of motion:${{v}^{2}}-{{u}^{2}}=2as$

$\Rightarrow {{(0)}^{2}}-{{(25)}^{2}}=2\times (-0.5)\times s$

$\Rightarrow -625=-s$

$\Rightarrow s=625m$

Therefore, the train covers a distance of $625m$ before it comes to rest.

19. A trolley, while going down an inclined plane, has an acceleration of$~2cm\text{/}{{s}^{2}}$. What will be its velocity $3s$ after the start?

Ans: It is given that,

Initial velocity of the trolley, $u=0$ (Trolley is starting from rest)

Acceleration of the trolley, $a=2cm/{{s}^{2}}=0.02m/{{s}^{2}}$

Time taken, $t=3s$

Final velocity (after $3s$ of start) of the trolley, $v=?$

It is known that,

From, first equation of motion: $v=u+at$

$\Rightarrow v=0+0.02(3)$

$\Rightarrow v=0.06m/s$

Thus, the velocity of the trolley is $0.06m/s$ after $3s$ from the start.

20. A racing car has a uniform acceleration of $4m{{s}^{-2}}$. What distance will it cover in $10s$ after start?

Ans: It is given that,

Initial velocity of the racing car, $u=0$ (The racing car is initially at rest)

Acceleration of a racing car, $a=4m/{{s}^{2}}$

Time taken, $t=10s$

It is known that,

From, second equation of motion: $s=ut+\frac{1}{2}a{{t}^{2}}$

$\Rightarrow s=0+\frac{1}{2}(4){{(10)}^{2}}$

$\Rightarrow s=200m$

Therefore, the distance covered by racing car after $10s$ from start is $200m$.

21. A stone is thrown in a vertically upward direction with a velocity of $5m\text{/}s$. If the acceleration of the stone during its motion is $10m/{{s}^{-2}}$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Ans: It is given that,

Initial velocity of the stone, $u=5m/s$

Final velocity, $v=0$ (Stone comes to rest after reaching maximum height)

Acceleration of the stone is equal to acceleration due to gravity, $a=-10m/{{s}^{2}}$ (Negative sign because of downward direction)

Maximum height reached by the stone, $s=?$

It is known that,

From, first equation of motion: $v=u+at$

$\Rightarrow 0=5+(-10)t$

$\Rightarrow 5=10t$

$\Rightarrow t=0.5s$

From, third equation of motion: ${{v}^{2}}-{{u}^{2}}=2as$

$\Rightarrow {{(0)}^{2}}-{{(5)}^{2}}=2(-10)s$

$\Rightarrow -25=-20s$

$\Rightarrow s=1.25m$

Therefore, the height attained by the stone is $1.25m$ in $0.5s$.

22. An athlete completes one round of a circular track of diameter $200m$ in $40s$. What will be the distance covered and the displacement at the end of $2$ minutes $20s$?

Ans: It is given that,

Diameter of a circular track, $d=200m$

Radius of the circular track, $r=\frac{d}{2}$

$\Rightarrow r=\frac{200}{2}=100m$

Circumference of the circular track, $c=2\pi r$

$\Rightarrow c=2\pi (100)=200\pi m$

Time taken to cover one round, $t=40s$

It is known that,

$Speed=\frac{Distance\text{ travelled}}{Time\text{ }taken}$

$\Rightarrow Speed=\frac{200\pi }{40}$

$\Rightarrow Speed=50\pi$

Athlete runs for $2$minutes $20$s: Time in seconds$=120+20=140s$

Total distance covered in $140s=Speed\times Time$

$\Rightarrow ~Distance=\frac{200\times 22\times 140}{40\times 7}=2200m$

Number of rounds$=\frac{140}{40}=3.5$

Athlete will be diametrically opposite to the point where he started after completing three rounds.

The displacement will be equal to diameter i.e.,$200m$

Therefore, the distance covered is $2200m$ and the displacement is $200m$at the end of $2$minutes $20$s.

23. Joseph jogs from one end A to the other end B of a straight road of $300m$ in $2$ minutes $50$ seconds and then turns around and jogs $100m$ back to point C in another $1$ minute. What are Joseph’s average speeds and velocities in jogging

1. from A to B

Ans: It is given that,

Distance from A to B$=300m$

Time taken from A to B $=2\min 50\sec =170\sec$

It is known that,

$Average\text{ }speed=\frac{Total\text{ }distance\text{ }covered}{Total\text{ }time\text{ }taken}$

$\Rightarrow Average\text{ }speed=\frac{300}{170}=1.765m/s$

$Average\text{ }velocity=\frac{Displacement}{Total\text{ }time\text{ }taken}$

Displacement from A to B$=\mathbf{Distance}=300m$

$Average\text{ }velocity=\frac{300}{170}=1.765m/s$

Therefore, the average speed and average velocity of Joseph from A to B are same and is equal to $1.765m/s$.

1. from A to C?

Ans: It is given that,

Distance from A to B$=300m$

Distance from B to C$=100m$

Total distance from A to C$=300+100=400m$

Time taken from A to B$=2\min 50\sec =170\sec$

Time taken from B to C$=1\min =60\sec$

Total time taken from A to C$=170+60=230\sec$

It is known that,

$Average\text{ }speed=\frac{Total\text{ }distance\text{ }covered}{Total\text{ }time\text{ }taken}$

$\Rightarrow Average\text{ }speed=\frac{400}{230}=1.739m/s$

$Average\text{ }velocity=\frac{Displacement}{Total\text{ }time\text{ }taken}$

Displacement from A to C$=AB-BC=300-100=200m$

$Average\text{ }velocity=\frac{200}{230}=0.87m/s$

Therefore, the average speed and average velocity of Joseph from A to C are $1.739m/s$ and $0.87m/s$ respectively.

24. Abdul, while driving to school, computes the average speed for his trip to be $20km{{h}^{-1}}$. On his return trip along the same route, there is less traffic and the average speed is $40km{{h}^{-1}}$. What is the average speed for Abdul’s trip?

Ans: It is given that,

Average speed of Abdul’s trip$=20km/h$

Let the distance travelled by Abdul to reach school and to return home be $d$.

Case 1: While driving to school

Let, total time taken be ${{t}_{1}}$.

$Average\text{ }speed=\frac{Total\text{ }distance\text{ }covered}{Total\text{ }time\text{ }taken}$

$\Rightarrow 20=\frac{d}{{{t}_{1}}}$

$\Rightarrow {{t}_{1}}=\frac{d}{20}$   ……$(1)$

Case 2: While returning from school

Let, total time taken be ${{t}_{2}}$.

$Average\text{ }speed=\frac{Total\text{ }distance\text{ }covered}{Total\text{ }time\text{ }taken}$

$\Rightarrow 40=\frac{d}{{{t}_{2}}}$

$\Rightarrow {{t}_{2}}=\frac{d}{40}$   ……$(2)$

Average speed for Abdul’s trip$=\frac{Total\text{ }distance\text{ }covered\text{ }in\text{ }the\text{ }trip}{Total\text{ }time\text{ }taken}$

Where,

Total distance covered in the trip$=d+d=2d$

Total time taken$={{t}_{1}}+{{t}_{2}}$

Substitute equation $(1)$ and $(2)$ in total time taken

Total time taken$=\frac{d}{20}+\frac{d}{40}$

$\Rightarrow Average\text{ }speed=\frac{2d}{\frac{d}{20}+\frac{d}{40}}$

$\Rightarrow Average\text{ }speed=\frac{2}{\frac{2+1}{40}}=\frac{80}{3}$

$\Rightarrow Average\text{ }speed=26.67m/s$

Therefore, the average speed for Abdul’s trip is $26.67m/s$.

25. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of $3.0m/{{s}^{2}}$ for $8.0s$. How far does the boat travel during this time?

Ans: It is given that,

The initial velocity of the motorboat, $u=0$( Motorboat is initially at rest)

Acceleration of the motorboat, $a=3m/{{s}^{2}}$

Time taken, $t=8s$

Distance travelled by the motorboat, $s=?$

It is known that,

From, second equation of motion: $s=ut+\frac{1}{2}a{{t}^{2}}$

$\Rightarrow s=0+\frac{1}{2}(3){{(8)}^{2}}$

$\Rightarrow s=(3)(8)(4)$

$\Rightarrow s=96m$

Therefore, the boat travels a distance of $96m$.

26. A driver of a car travelling at $52km{{h}^{-1}}$ applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at $3km{{h}^{-1}}$ in another car applies his brakes slowly and stops in $10s$. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Ans: Car A: Initial speed of the car, ${{u}_{A}}=52km/h$

$\Rightarrow {{u}_{A}}=52\times \frac{5}{18}=14.4m/s$

Time taken for the car to stop, ${{t}_{A}}=5s$

Final speed of the car becomes zero after $5s$ application of brakes.

Car B: Initial speed of the car, ${{u}_{B}}=3km/h$

$\Rightarrow {{u}_{B}}=3\times \frac{5}{18}=0.833m/s$

Time taken for the car to stop, ${{t}_{B}}=10s$

Final speed of the car becomes zero after $10s$ application of brakes.

Plot of the two cars on a speed-time graph is shown below:

Distance covered by each car is equal to the area under the speed-time graph.

Distance covered by car A: ${{s}_{A}}=\frac{1}{2}\times OP\times OR$

$\Rightarrow {{s}_{A}}=\frac{1}{2}\times 14.4\times 5$

$\Rightarrow {{s}_{A}}=36m$

Distance covered by car B: ${{s}_{B}}=\frac{1}{2}\times OS\times OQ$

$\Rightarrow {{s}_{B}}=\frac{1}{2}\times 0.83\times 10$

$\Rightarrow {{s}_{B}}=4.15m$

Area of $\Delta OPR$is greater than area of $\Delta OSQ$.

Therefore, the distance covered by car A is greater than the distance covered by car B. Thus, the car travelling with a speed of $52km/h$travels farther after the brakes were applied.

27. The following figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following:

1. Which of the three is travelling the fastest?

Ans: It is known that,

$Speed=\frac{Distance}{Time}$

$\mathbf{Slope}\text{ }\mathbf{of}\text{ }\mathbf{graph}=\frac{y-axis}{x-axis}=\frac{Distance}{Time}=Speed$

Slope of the graph of object B is greater than objects A and C.

Therefore, object B is travelling the fastest.

1. Are all three ever at the same point on the road?

Ans: No, All the three objects A, B and C never meet at the same point.

Therefore, they were never at the same point on the road.

1. How far has C travelled when B passes A?

Ans: From the graph,

On distance axis:$7$small boxes $=4km$

$\Rightarrow \mathbf{1}\text{ }\mathbf{small}\text{ }\mathbf{box}=\frac{4}{7}km$

Initially, object C is $4$ blocks away from the origin$\Rightarrow \frac{16}{7}km$

Distance of object C from origin when B passes A is $8km$.

Distance covered by C$=8-\frac{16}{7}=5.714km$

Therefore, C has travelled a distance of $5.714km$ when B passes A.

1. How far has B travelled by the time it passes C?

Ans: From the graph,

Distance covered by B at the time it passes C$=9boxes$

Distance$=9\times \frac{4}{7}=\frac{36}{7}=5.143km$

Therefore, B has travelled a distance of $5.143km$ when B passes A.

28. A ball is gently dropped from a height of $20m$. If its velocity increases uniformly at the rate of $10m/{{s}^{2}}$, with what velocity will it strike the ground? After what time will it strike the ground?

Ans: It is given that,

Distance covered by the ball, $s=20m$

Acceleration of the ball, $a=10m/{{s}^{2}}$

Initial velocity of the ball, $u=0$ (Ball is initially at rest)

Final velocity of the ball, $v=?$

Time taken by the ball to strike ground, $t=?$

It is known that,

From, third equation of motion: ${{v}^{2}}-{{u}^{2}}=2as$

$\Rightarrow {{(v)}^{2}}-{{(0)}^{2}}=2(10)(20)$

$\Rightarrow {{v}^{2}}=400$

$\Rightarrow v=20m/s$

From, first equation of motion: $v=u+at$

$\Rightarrow 20=0+(10)t$

$\Rightarrow 20=10t$

$\Rightarrow t=2s$

Therefore, the ball strikes the ground after $2s$ with a velocity of $20m/s$.

29. The speed-time graph for a car is shown as a figure.

1. Find out how far the car travels in the first $4$seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

Ans: Distance travelled by the car during the first $4$seconds is equal to the area of the shaded region on the graph.

$Area=\frac{1}{2}\times 4\times 6=12m$

Therefore, the distance travelled by the car in first $4$seconds is $12m$.

1. Which part of the graph represents uniform motion of the car?

Ans: Horizontal line after $6$seconds represents the constant motion.

Therefore, the shaded part of the graph between time $6$seconds to $10$ seconds represent the uniform motion of the car.

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

1. An object with a constant acceleration but with zero velocity.

Ans: Possible.

For example, when a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, equal to $9.8m/{{s}^{2}}$.

1. An object moving in a certain direction with an acceleration in the perpendicular direction.

Ans: Possible.

For example, when an object is moving in a circular track, its acceleration is perpendicular to the direction of velocity.

## CBSE NCERT Solutions for Class 9 Science: Motion – A Brief Overview

You can opt for Chapter 8 - Motion NCERT Solutions for Class 9 Science PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Science

Along with this, students can also view additional study materials provided by Vedantu, for Class 9

According to experts, everything in this universe is constantly moving. This means that even if you are completely still, then you are still standing or sitting on earth that is constantly in motion around the sun and its axis.

Motion can be defined as any change that is in the position of an object about time. In physics NCERT solutions class 9 science chapter 8, motion can be explained as the change in the position of an object in respect to its surroundings in any given interval of time.

It is also mentioned in the class 9 chapter 8 science book that the motion of an object, which has some mass, can be expressed in terms of displacement, distance, time, acceleration, time, velocity, and speed.

There are many examples of motion, including a book that is falling from a table, the flow of water from a tap, rattling of windows, and the rapid flapping of a flag. These examples can be mentioned in class 9 science chapter 8 solutions.

It should be noted that even the air that we breathe exhibits motion. The fundamental particles of matter that are present in everything are also in a state of constant motion. It is also possible for motion to be either slow or swift.

Readers might also be interested to learn that displacement can be defined as the shortest distance between the final and initial position. These definitions and concepts can come in handy to solve NCERT class 9 science chapter 8 questions.

### There Are Also Other Examples of Motion. Some of Those Examples Are

• The flow of air around everybody and the flow of air in and out of lungs

• Various daily activities like running, walking, and closing the door

• The vehicles that carry individuals or passengers from one location to another location. In this case, there is a change in the position of the passengers from one place to another

Students should be familiar with all the basic definitions and examples of motion. This information can be used for writing NCERT class 9 science chapter 8 solutions.

## The Types of Motion

According to various class 9 science ch 8 solutions, the motion of an object is directly related to the type of force that is acting on the body. This means that there can be different types of motions. Do you know what those different types of motions are? If you don’t, then go through the list that is mentioned below.

• Translational Motion

Translational is a type of motion in which the object moves along a path that is carved out in any of three dimensions.

• Rotational Motion

Rotational is a type of motion in which the object is moved along a circular path. This object has a fixed axis. It is also known as rotatory motion. Some examples of rotational or rotatory motion are:

1. The motion of the earth around its axis and the sun

2. The motion of the wheels and steering wheel about its axis. This motion can be seen when one is driving a car

• Linear Motion

Linear is a type of translational motion. In this type of motion, the body moves along a single direction in a single dimension. It can also be defined as the type of motion in which the body moves from one point to another in a curved path or straight line.

There are also two different types of linear motions based on the path of the motion. And those types of linear motions are:

1. Rectilinear Motion: The path of this motion is along a straight line

2. Curvilinear Motion: The path of motion is curved

According to the ch 8 class 8 science book, some examples of linear motion are the motions of a football, train, and a car running on a road.

• Periodic Motion

Periodic motion is a type of motion in which the motion repeats itself after certain intervals of time.

• Simple Harmonic Motion

Simple harmonic motion is similar to the motion of a simple pendulum. In this motion, there is a restoring force that acts in the direction that is opposite to the direction of motion of the object.

The value of the restoring force is also proportional to the displacement of the object from its mean position. Students should be familiar with this type of motion if they want to write NCERT solutions for class 9 science motion chapter correctly.

• Projectile Motion

In this type of motion, there is a horizontal displacement as well as vertical displacement.

• Oscillatory Motion

Oscillatory motion is repetitive, and it exists within a time frame. If the oscillatory motion is mechanical, then it is known as vibration. This motion exists around the mean position of a body.

Students should also be familiar with examples of oscillatory motion if they want to answer all chapter 8 science class 9 NCERT solutions correctly. Some of those examples are mentioned below.

• When a child is playing on a swing and is pushed. The swing moves in a to and fro motion about its mean position

• The pendulum of a clock that moves to and fro about its mean position

• The strings of the guitar, when the instrument is strummed, moves in a to and fro motion about its mean position

### The Laws of Motion

There are many questions related to laws of motion in the NCERT book. Students often face challenges in answering those questions. This is where the class 9 science chapter 8 NCERT solutions pdf might be of help.

Students can download NCERT class 9 science ch 8 solutions from Vedantu for free. The class 9 NCERT science chapter 8 solutions are written by the most talented experts in India and can help students to score good marks.

For now, let’s focus on understanding the laws of motion given by Newton. Newton’s laws of motion are mentioned below.

• First Law

The first law states that any object that remains in its existing state of motion or rest unless a net external force is applied on the object.

• Second Law

The second law states that an object that has a certain mass, the greater the mass of the object, the greater will be the force that will be required for accelerating the object. This law can be depicted by the equation F = ma.

In this equation, ‘F’ is the force on the object, ‘m’ is the mass of the object, and ‘a’ is the acceleration of the object.

• Third Law

According to the third law of motion, for every action, there is an equal and opposite reaction.

It should also be noted that there are some limitations to these laws. However, these laws still hold in the current world.

Students should also learn the NCERT solutions for class 9 science ch 8 to score good marks. Students can download NCERT Solutions for class 9 chapter 8 science pdf from Vedantu.

The NCERT Solutions for Motions, Chapter 8 is required to be studied by the students while they prepare for their Class 9 CBSE exam. We also have covered the discussion on important topics from this chapter - Motion.

Students must also take care of the numerical present in this chapter, and revise the formulae regularly.

## FAQs on Motion - NCERT Solutions of Chapter 8 (Science) for Class 9

1. In regards to the NCERT Solutions of Chapter 8, Can an object have zero displacements if it has moved through a distance? If yes, then support your answer with an example or proof.

Yes, an object can move a certain distance without having any displacement. The logic behind this statement is that the displacement is defined as the shortest distance between the final and initial position of the object.

This means that even if an object moves through a large distance, it should come back to its initial position. In that case, the corresponding displacement of the object will be zero.

2. Mention the condition under which the magnitude of the average velocity of an object can be equal to the value of average speed.

We know that the average speed is the total distance that is travelled in a time frame and velocity is the total displacement in the time frame. The magnitude of average velocity and average speed will be the same when the total distance that is travelled is equal in its value to the displacement.

3. What will I learn from Chapter 8 of NCERT Solutions for Class 9 Science?

Chapter 8 of Class 9 Science is “Motion”. Students can learn about motion, its examples, types of motion including translational motion, rotational motion, linear motion, periodic motion, simple harmonic motion, projectile motion and oscillatory motion, and the laws of motion, from the NCERT Solutions for Chapter 8 of Class 9 Science. They can refer to the solutions to easily understand the concepts, especially if they have any confusion.

4. How can I solve the problems from the NCERT Solutions for Class 9 Science Chapter 8 effortlessly?

Students should try to solve all the questions and numerical problems of Chapter 8 of the Class 9 Science NCERT textbook. If they get stuck on any question or have a doubt about any topic, they can refer to the NCERT Solutions for Class 9 Science Chapter 8.  The answers to all questions from the NCERT textbook of Class 9 Science are provided in the NCERT Solutions.

5. Is it important to study NCERT Solutions for Class 9 of Chapter 8?

Yes, ut is very important for the students to study NCERT solution of every chapter to score good marks in CBSE Class 9 exam. By studying the NCERT study material they will exactly know from which portion to study so that they can perform well in their exam.

6. What is motion and types of motion in the NCERT Solutions of Class 9?

Any change in the position of an object is defined as motion. There are various types of motion, including translational motion, rotational motion, linear motion (rectilinear motion and curvilinear motion), periodic motion, simple harmonic motion, projectile motion, and oscillatory motion. This is because the motion of an object is directly related to the type of force acting on the body.

7. How to download the NCERT Solutions for Class 9 Science Chapter 8?

Students can download by clicking on NCERT Solutions for Class 9 Science Chapter 8. On this link, you will find the NCERT Solutions for Chapter 8 of Class 9 Science provided by Vedantu free of cost. You can also find the NCERT Solutions for all chapters of Class 9 Science as well as other subjects such as Maths, English, Hindi, and Social Science. All these can be downloaded from the Vedantu app also. Several other study materials such as important questions, revision notes and previous year question papers are also available on Vedantu (vedantu.com).