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NCERT Solutions for Class 9 Science Chapter 7 Motion

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NCERT Solutions for Class 9 Science Chapter 7 - Motion FREE PDF Download

The updated NCERT Solutions for Chapter 7 Motion Class 9 is now available on Vedantu. Our subject experts prepare these solutions, referencing the latest NCERT Class 9 Science textbook edition.

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Students can download the NCERT Solutions for Class 9 Science PDF and refer to these solutions for FREE from our website. These solutions have included all the important topics and sub-topics, such as motion, velocity, and uniform circular motion, covered in Class 9 science chapter 7, according to the latest Class 9 CBSE Science Syllabus. Therefore, students can rely upon these NCERT Solutions to prepare for their exams.

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NCERT Solutions for Class 9 Science Chapter 7 Motion
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Access Science Class 9 Motion Chapter 7 NCERT Solutions

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.


A man is walking in a square park


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

  1. The distance travelled by an object in a given interval of time is speed.

  2. Speed does not have any direction.

  3. Speed is either positive or zero but not negative.

  1. The displacement of an object in a given interval of time is velocity.

  2. Velocity has a unique direction.

  3. 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.


Distance-time graph for uniform motion


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


Distance-time graph for non-uniform motion


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.


Distance-time graph for stationary object


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.


Speed time graph of a body moving unifromly


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.


Area of velocity time graph


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 \[=2\min 50\sec =170\sec\]


Motion of Joseph while jogging


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$=300+100=400m$


Complete motion of Joseph while jogging


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:


Speed time graph of two cars


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:


Distance-time graph of three objects


  1. Which of the three is travelling the fastest? 

Ans: It is known that,

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


Distance time graph of three objects


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

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 $=4km$

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


Determination of distance travelled from the graph


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$=9boxes$


Determination of distance travelled by object B from the graph


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.


Speed-time graph for a car


  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.


Distance travelled by the car


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

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.


Speed time graph of the car


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.



31. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Ans. Given, the radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2π42250km = 265571.42 km

Time is taken for the orbit = 24 hours

Therefore, speed of the satellite = 11065.4 km.h-1

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.


Topics Covered in Class 9 Science Chapter 7 - Motion

List of Topics Covered in Class 9 Science Chapter 7 - Motion

Topics

Subtopics

Motion

Motion Along a Straight Line, Uniform and Non-Uniform Motion, Speed with Direction,

Rate of Change of Velocity


Measuring the Rate of Motion


Graphical Representation of Motion

Distance-time graphs, Velocity-time graphs, 

Equations of Motions


Uniform Circulation Motion




Deleted Topics in Class 9 Science Chapter 7 Motion

  • Equations of Motion by Graphical Method

  • Equation for Velocity-Time Relation

  • Equation for Position-Time Relation

  • The equation for Position Velocity


Benefits of NCERT Solutions for Class 9 Science Chapter 7 Motion

  • NCERT Solutions motion class 9 questions answers uses a simple and easy-to-understand approach to teach students about various topics.

  • All the questions in the relevant NCERT textbooks are entirely solved for class 9 science chapter 7 question answer.

  • To assist students in their preparations, NCERT Solutions provides complete answers to NCERT solutions for class 9 science chapter 7 motion exercise.

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

  • 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.

  • 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.


Important Study Material Links for Class 9 Science Chapter 7 Motion

S.No. 

Important Links for Class 9 Chapter 7 Science Study Material 

1.

Class 9: Motion Important Questions

2.

Class 9: Motion Revision Notes

3.

Class 9: Motion Exemplar Solutions



Conclusion

Motion class 9 NCERT solutions provide students with simple and detailed definitions and explanations of each concept covered in the chapter. Therefore, it is highly recommended that students download and refer to our comprehensive and expert-curated class 9 science ch 7 NCERT Solutions to get a gist of the chapter before the exam and to know how to answer the questions in the exam. Students can also refer to our plethora of other study resources related to this chapter, which are available for free on our website.


NCERT Solutions for Class 9 Science - Other Chapter-Wise Links

Given below are the links for the other chapter-wise NCERT Solutions for Class 9 Science. These solutions are provided by the Science experts at Vedantu in a detailed manner. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Study Material Links for Class 9 Science

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FAQs on NCERT Solutions for Class 9 Science Chapter 7 Motion

1. Why are the NCERT Solutions for Class 9 Science Chapter 7 important for the 2025-26 exams?

The NCERT Solutions for Class 9 Science Chapter 7 are crucial as they provide the authoritative, step-by-step methodology for solving all in-text and exercise questions as per the latest CBSE 2025–26 syllabus. They help students understand the correct format for writing answers, including formulas, unit conversions, and final statements, which is essential for scoring full marks.

2. What is the best way to use the NCERT Solutions for solving problems in Chapter 7, Motion?

For effective learning, first attempt to solve the NCERT textbook problems on your own. Then, use the provided solutions to verify your answers and, more importantly, to understand the step-by-step method. Pay close attention to how concepts like acceleration, velocity, and displacement are applied, and analyse the steps where your method differed. This approach builds problem-solving skills rather than encouraging rote memorization.

3. How do the NCERT Solutions explain the concept of zero displacement even when an object has travelled a distance?

The NCERT solutions explain this by breaking down the definitions. They clarify that distance is the total path length covered, while displacement is the shortest straight-line distance between the initial and final points. The solution for the example of an object moving in a circular path and returning to the start shows that while the distance is the circumference (2πr), the displacement is zero because the initial and final positions are the same.

4. How are distance-time and velocity-time graphs used to solve numericals in the NCERT Class 9 Chapter 7 solutions?

The solutions demonstrate that graphs are powerful tools for solving motion problems. They show how to:

  • Calculate an object's speed from the slope of a distance-time graph.
  • Determine the distance travelled by calculating the area under a velocity-time graph.
  • Identify the nature of motion (uniform, non-uniform, at rest) by interpreting the shape of the graph, such as a straight line or a curve.

5. How do the NCERT Solutions for Chapter 7 differentiate between speed and velocity?

The NCERT solutions clearly distinguish between speed and velocity by highlighting two key differences:

  • Nature of Quantity: Speed is a scalar quantity, meaning it only has magnitude (e.g., 50 km/h). Velocity is a vector quantity, possessing both magnitude and direction (e.g., 50 km/h North).
  • Calculation: Speed is calculated as total distance divided by time, whereas velocity is calculated as displacement divided by time. Because of this, velocity can be zero even when speed is not.

6. How should I correctly apply the sign conventions when solving numericals on acceleration from NCERT Chapter 7?

The NCERT solutions demonstrate a crucial rule: the sign depends on the direction of change in velocity. If an object's velocity is increasing in the direction of motion, the acceleration is positive. If the velocity is decreasing (like when brakes are applied), the acceleration is in the opposite direction and is considered negative. This negative acceleration is also known as retardation or deceleration, a key concept for solving problems accurately.

7. Many students confuse average speed and average velocity. How do the solved problems in NCERT Solutions for Chapter 7 (like the 'Abdul's trip' problem) clarify this difference?

The 'Abdul's trip' problem is a perfect illustration. The solution shows that to find average speed, you must calculate the total distance (to school and back) and divide it by the total time taken. However, to find average velocity for the entire trip, the displacement is zero (since Abdul returns to the starting point), making the average velocity zero. This example powerfully demonstrates that average speed depends on the entire path, while average velocity only depends on the start and end points.

8. The NCERT solutions for Chapter 7 include problems on uniform and non-uniform acceleration. Why is understanding the solution method for both types important?

Understanding both is vital because they describe different, realistic motion scenarios. The solutions for uniform acceleration problems use the three equations of motion (e.g., v = u + at), which apply only when acceleration is constant. In contrast, problems involving non-uniform acceleration, often represented by curved graphs, require a conceptual understanding that velocity changes by unequal amounts in equal time intervals. The solutions teach you to first identify the type of acceleration before choosing the correct formula or graphical method.

9. When solving problems involving the three equations of motion, what is a common mistake and how do the NCERT Solutions help prevent it?

A very common mistake is incorrect unit conversion. For example, students often forget to convert speed from km/h to m/s before using it in equations where time is in seconds. The NCERT solutions consistently show the conversion step (e.g., multiplying km/h by 5/18 to get m/s) right at the beginning of the solution. Following this structured approach, as shown in the solutions, helps build a habit of checking units first, preventing major errors in the final answer.

10. The NCERT solutions cover both linear and circular motion problems. What fundamental principle connects the solutions for a car moving on a straight road and an artificial satellite in orbit?

The connecting principle is the concept of velocity. For the car on a straight road, a change in velocity often means a change in speed. For the satellite in a circular orbit, the solutions highlight a more subtle point: even if the speed is constant, the direction of motion is continuously changing. This change in direction means the velocity is changing, which implies the satellite is constantly accelerating towards the centre of the orbit. The solutions thus teach that acceleration isn't just about speeding up or slowing down.