## CBSE Class 9 Science Chapter-7 Important Questions - Free PDF Download

Class 9 plays a vital role in the lives of the students. It is the foundation class for the upcoming Board examinations of class 10th. The curriculum and entire academic marking of class 9th are also comparatively more challenging, and the students must learn the subjects well to score fantastic marks in the examinations. Motion chapter in **Class 9th Science** involves various concepts related to Velocity, distance, and Displacement, requiring more preparation and practice.

**Important questions for Class 9th Science Chapter 7** are essential to help the students get a brief practice during the exam times and score well in the examinations. Vedantu provides precisely designed **Class 9th Science Motion Important Questions**, based on the previous year papers' research. The expert team prepares the study materials to ensure the best practice resource for the students.

Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. You can download Class 9 Maths and Class 9 Science NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

## Download CBSE Class 9 Science Important Questions 2024-25 PDF

Also, check CBSE Class 9 Science Important Questions for other chapters:

CBSE Class 9 Science Important Questions | ||

Sl.No | Chapter No | Chapter Name |

1 | Chapter 1 | |

2 | Chapter 2 | |

3 | Chapter 3 | |

4 | Chapter 4 | |

5 | Chapter 5 | |

6 | Chapter 6 | |

7 | Chapter 7 | |

8 | Chapter 8 | Motion |

9 | Chapter 9 | |

10 | Chapter 10 | |

11 | Chapter 11 | |

12 | Chapter 12 | |

13 | Chapter 13 | |

14 | Chapter 14 | |

15 | Chapter 15 |

Important Questions for CBSE Class 9 Science Chapter 7 - Motion

## Study Important Questions for Class 9 Science Chapter 7 - Motion

Very Short Answer Questions (1 Mark)

1. Which of the following statements is correct?

Both speed and velocity are same

Speed is a scalar and velocity is a vector

Speed is a vector and velocity is scalar

None of these

Ans: b) speed is a scalar and velocity is a vector

2. What is the slope of the body when it moves with uniform velocity?

Positive

Negative

Zero

May be positive or negative

Ans: b)Zero

3. Which of the following is the position time graph for a body at rest?

a)

b)

c)

d)

Ans: a)

4. What does an area velocity time graph give?

Distance

Acceleration

Displacement

None of the above

Ans: c) Displacement

5. If a body starts from rest, what can be said about the acceleration of the body?

Positively accelerated

Negative accelerated

Uniform accelerated

None of the above

Ans: a) Positively accelerated

6. What does the slope of the position-time graph give?

Speed

Acceleration

Uniform speed

Both (a) and (c) depending upon the type of graph.

Ans: a) Speed

7. When a body moves uniformly along the circle, then:

Its velocity changes but speed remain the same

Its speed changes but velocity remain the same

Both speed and velocity changes

Both speed and velocity remains the same

Ans: a) Its velocity changes but speed remains the same

8. Which of the following statements is correct?

Speed distance are scalar, velocity and displacement are vector

Speed distance are vector, velocity and displacement are vector

Speed and velocity are scalar, distance and velocity are vector

Speed and velocity are vector, distance and displacement are scalar

Ans: a) Speed distance are scalar, velocity and displacement are vector

9. What does the slope of a velocity-time graph give?

Distance

Displacement

Acceleration

Change in velocity.

Ans: c) Acceleration

10. The displacement of the body can be-

Positive

Negative

Zero

All of these.

Ans: d) All of these.

11. Which of the following gives both direction and magnitude-

Scalar

Vector

Both

None.

Ans: b) Vector

12. If a moving body comes to rest, then its acceleration is

Positive

Negative

Zero

All of these depending upon initial velocity.

Ans: b) Negative

Short Answer Questions (2 Marks)

1. Distinguish between speed and velocity.

Ans: Speed of a body is the distance travelled by a body as per unit time while velocity is the rate and direction of an objectโs movement.

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

Ans: If the distance travelled by a body is equal to the displacement, then the magnitude of the average velocity of an object will be equal to its average speed.

3. What does the odometer of an automobile measure?

Ans: The odometer of an automobile is used to measure the distance covered by an automobile.

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

Ans: Graphically the path of an object will be linear; it looks like a straight line when it is in uniform motion.

5. 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, \[\mathbf{3}\times \mathbf{108m}{{s}^{-1}}\]

Ans: The given data is that time is five minutes and speed is$(3\times {{10}^{8}}m{{s}^{-1}})$

\[Distance=Speed\times Time\]

$\Rightarrow 5\min \times (3\times {{10}^{8}}m{{s}^{-1}})$

$\Rightarrow (5\times 60)\sec \times (3\times {{10}^{8}}m{{s}^{-1}})$

$\Rightarrow 300\sec \times (3\times {{10}^{8}}m{{s}^{-1}})$

$\Rightarrow 900\times {{10}^{8}}m{{s}^{-1}}=9\times {{10}^{10}}m$

$\therefore Dis\tan ce=9\times {{10}^{7}}km$

6. Which of the following is true for displacement?

It cannot be zero.

Ans: False with respect to the concept of displacement

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

Ans: False with respect to the concept of displacement.

7. When will you say a body is in

Uniform acceleration?

Ans: When an object travels in a straight line and its velocity changes by equal amount in an equal interval of time, it is said to have uniform acceleration.

Non-uniform acceleration?

Ans: Non-uniform acceleration is also called variable acceleration. When the velocity of an object changes by unequal amounts in equal intervals of time, it is said to have non-uniform acceleration.

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

Ans: Initial speed of bus (u) $=80km{{h}^{-1}}$

$\Rightarrow \dfrac{80\times 1000}{60\times 60\sec }$

$\Rightarrow \dfrac{200}{9m{{s}^{-1}}}=22.22m{{s}^{-1}}$ Final speed of bus (v) $=60km{{h}^{-1}}$

$\Rightarrow \dfrac{60\times 1000}{60\times 60\sec }=\dfrac{50}{3m{{s}^{-1}}}=16.67m{{s}^{-1}}time(t)=5s$

Acceleration (a) $=\dfrac{(v-u)}{t}=\dfrac{(16.67-22.22)}{5}=\dfrac{-5.55}{5}=-1.11m/{{s}^{2}}$

$\therefore Acceleration\text{ }\left( a \right)=-1.11m/{{s}^{2}}$

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

Ans: If an object has a uniform motion then the nature of distance time graph will be linear, that is it would in a straight line and if it has non-uniform motion then the nature of the distance-time graph will be a curved line.

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

Ans: The area occupied below the velocity-time graph measures the distance moved by any object.

11. A bus starting from rest moves with a uniform acceleration of \[\mathbf{0}.\mathbf{1m}{{\mathbf{s}}^{-2}}\] for \[\mathbf{2}\text{ }\mathbf{minutes}\] . Find

The speed acquired,

Ans: $u=0,a=0.1m{{s}^{-2}},t=2\min =120\sec $

$v=u+at=0+0.1\times 120=12m{{s}^{-1}}$

\[\mathbf{Speed}\text{ }\mathbf{acquired}=v=12m{{s}^{-1}}\]

The distance travelled.

Ans: $s=ut+\dfrac{1}{2}a{{t}^{2}}=0\times 120+\dfrac{1}{2}0.1\times {{120}^{2}}=720m$

12. A trolley, while going down an inclined plane, has an acceleration of$2cm{{s}^{-2}}$ . What will be its velocity \[\mathbf{3s}\]after the start?

Ans: Given:$u=0,a=2cm/{{s}^{2}},t=3s$

$v=u+at=0+2\times 3=6cm/s$

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

Ans: Given:$u=0,a=4m/{{s}^{2}},t=10s$

$s=ut+\dfrac{1}{2}a{{t}^{2}}$

$s==0\times 10+\dfrac{1}{2}\times 4\times {{10}^{2}}$

$\therefore s=200m$

14. Differentiate between distance and displacement?

Ans: The difference between distance and displacement is as below,

Distance | Displacement | |

1 | The length of the actual path travelled by the body from initial position to final position | The length of the straight line joining the initial and final position of the body |

2 | It is a scalar quantity, that is it has only magnitude | It is a vector quantity that is it has both magnitude and direction. |

3 | It is always positive. | It may be positive, negative or zero. |

15. Derive mathematically the first equation of motion\[\mathbf{V}=\mathbf{u}+\mathbf{at}\]?

Ans: Acceleration is defined as the rate of change of velocity.

Let V=final velocity; \[{{\mathbf{V}}_{o}}\]= initial velocity, T= time, a = acceleration.

So by definition of acceleration

$a=\dfrac{V-{{V}_{o}}}{T}$

$at=V-{{V}_{o}}$

$V={{V}_{o}}+at$

If \[{{V}_{o}}=u=\] initial velocity, then \[\left[ V=\text{ }u+at \right]\]

16. Calculate the acceleration of a body which starts from rest and travels \[\mathbf{87}.\mathbf{5m}\text{ }\mathbf{5sec}\]?

Ans: Given Data: $u=0$ (starts from rest) u= initial velocity

a=? a=acceleration

\[T=5\sec ,t=time\]

\[S=\text{ }87.5m\text{ }\left( S=distance \right)\]

From second equation of motion

$S=ut+\dfrac{1}{2}a{{t}^{2}}$

$\Rightarrow 87.5=0+\dfrac{1}{2}a{{t}^{2}}$

$\Rightarrow 87.5=\dfrac{1}{2}a{{t}^{2}}\to (i)$

$\Rightarrow 87.5\times 2=a\times {{(5)}^{2}}$

$\Rightarrow \dfrac{87.5\times 2}{25}=a$

$\Rightarrow \dfrac{175.0}{25}=a$

$\therefore S=7m/{{s}^{2}}=a$

17. Define uniform velocity and uniform acceleration?

Ans: Uniform velocity:- A body is said to move with uniform velocity if equal displacement takes place in equal intervals of time, however small these intervals may be.

Uniform acceleration:- A body is said to move with uniform acceleration if equal changes in velocity take place in equal intervals of time, however, small intervals may be.

18. A car travels at a speed of \[\mathbf{40km}/\mathbf{hr}\]for two hour and then at \[\mathbf{60km}/\mathbf{hr}\]for three hours. What is the average speed of the car during the entire journey?

Ans: Given: In first case;${{t}_{1}}=time=2hrs$

${{v}_{1}}=speed=40km/hr$

${{s}_{1}}=dis\tan ce=speed\times time$

${{s}_{1}}=40\times 2=80km$

In second case, Given ${{t}_{2}}=time=3hrs$

${{v}_{2}}=speed=60km/hr$

${{s}_{2}}=dis\tan ce=speed\times time$

${{s}_{2}}=60\times 3=180km$

The total distance = ${{s}_{1}}+{{s}_{2}}=80+180=260km$

Total time, ${{t}_{1}}+{{t}_{2}}=2+3=5hrs$

Average speed = $\dfrac{total\text{ }distance}{total\text{ }time}=\dfrac{260}{5}$

$\therefore Average\text{ }speed=52km/hr$ .

19. The velocity-time graph of two bodies A and B traveling along the +x direction are given in the figure

Are the bodies moving with uniform acceleration?

Ans: Yes the bodies are moving with uniform acceleration.

Which body is moving with greater acceleration A or B?

Ans: Body A is moving with greater acceleration.

20. Derive the second equation of motion, \[\mathbf{s}=~\mathbf{ut}+~\dfrac{1}{2}a{{t}^{2}}\] numerically?

Ans: Let at time \[\mathbf{t}=\mathbf{0}\] , body has initial velocity = ${{V}_{o}}$

At timeโtโ, the body has final velocity=V

S=distance traveled in timeโtโ

We know, total distance traveled = Average velocity $\times $ time

\[Average\text{ }velocity~~=\dfrac{initial\text{ }velocity+final\text{ }velocity}{2}\]

$\Rightarrow \dfrac{{{V}_{o}}+V}{2}$

$Total~distance=s=\dfrac{{{V}_{o}}+V}{2}\times t$

$\Rightarrow 2s=({{V}_{o}}+V)t\to (i)$

Now from first equation of motion, $V={{V}_{o}}+at\to (ii)$

Use the value of (V) from (ii) in (i)

$2s=({{V}_{o}}+{{V}_{o}}+at)t$

$2s=2{{V}_{o}}t+\dfrac{1}{2}a{{t}^{2}}$

$Let,{{V}_{o}}=u$

$\Rightarrow s=ut+\dfrac{1}{2}a{{t}^{2}}$

21. Calculate the acceleration and distance of the body moving with $5m/{{s}^{2}}$ which comes to rest after traveling for 6sec?

Ans: Acceleration=a=?

Final velocity = V =o (body comes to rest)

Distance = s =?

\[Time\text{ }=t=\text{ }6~sec\]

From, \[V~=~u~+~at\]

$O=5+a\times 6$

$-a\times 6=5$

$a=\dfrac{-5}{6}m/{{s}^{2}}$

Now,

${{v}^{2}}-{{u}^{2}}=2as$

${{O}^{2}}-25=2\times \dfrac{-5}{6}\times s$

$\Rightarrow -25=\dfrac{-5}{3}\times s$

$\Rightarrow \dfrac{25\times 3}{5}=s$

$\therefore s=15m$

22. A body is moving with a velocity of $12m/s$ and it comes to rest in 18m, what was the acceleration?

Ans: Initial velocity\[=u=12m/s\]

Find velocity =V=0

\[S=distance=18m\]

A= acceleration =?

From\[~{{3}^{rd}}\]equation of motion;

${{v}^{2}}-{{u}^{2}}=2as$

${{O}^{2}}-{{(12)}^{2}}=2\times a\times 18$

$\Rightarrow \dfrac{-144}{36}=a$

$a=\dfrac{-144}{36}$

$[a=-4m/{{s}^{2}}]$

$\Rightarrow 4m/{{s}^{2}}$

23. A body starts from rest and moves with a uniform acceleration of $4m/{{s}^{2}}$until it travels a distance of $800m$, find the find velocity?

Ans: Initial velocity $=u=0$

Final velocity$=v=?$

Acceleration$=a=4m/{{s}^{2}}$

Distance$=s=800m$

${{v}^{2}}-{{u}^{2}}=2as$

${{u}^{2}}-(0)=2\times 4\times 800$

$u=80m/s$

$\therefore {{u}^{2}}=6400$

24. The driver of a car traveling along a straight road with a speed of $72kmph$ observes a sign board which gives the speed limit to be$54kmph$. The signboard is $70m$ahead when the driver applies the brakes calculate the acceleration of the car which will cause the car to pass the signboard at the stated speed limit?

Ans: Initial speed \[=u=72\text{ }km/hr\]

$\Rightarrow \dfrac{72\times 5}{18}=20m/s$

Final speed \[=v=54\text{ }km/hr\]

$\Rightarrow \dfrac{54\times 5}{18}=15m/s$

Distance\[~~=~\mathbf{S}~=~\mathbf{70m}\]

Now, ${{v}^{2}}-{{u}^{2}}=2as$

${{(15)}^{2}}-{{(20)}^{2}}=2\times a\times 70$

$\Rightarrow 225-400=140a$

$\Rightarrow -175=140a$

$[a=-1.25m/{{s}^{2}}]$

25. Differentiate between scalars and vectors?

Ans: The difference between scalars and vectors is as below,

Vector | Scalar |

It has magnitude and specific direction | It has magnitude but no direction |

It can be positive or negative | It is always positive |

Eg. Displacement, velocity | Eg: Distance, speed |

Short Answer Questions (3 Marks)

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

Ans: Yes, if an object is moved through a distance it can have zero displacement because displacement of an object is the actual change in its position when it moves from one position to the other position. So if an object travels from point A to B and then returns back to point A again, the total displacement will be 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\min 20\sec $ ?

Ans: Distance covered by farmer in \[\mathbf{40}\text{ }\mathbf{seconds}~=4\times (10)m=40m\]

Speed of the farmer = distance/time \[=40m/40s=1m/s.\]

Total time given in the question\[=2min20sec=60+60+20=140sec\]

Since he completes $1round$of the field in \[40\]seconds so in he will complete $3$rounds in \[120\text{ }seconds~\left( 2mins \right)or120m\] distance is covered in\[2min\]. In another \[20sec\]will cover another\[20m\]so total distance covered in \[2min20sec=120+20=140m.\]

\[Displacement=\sqrt{{{10}^{2}}}+{{10}^{2}}\]

$\Rightarrow \sqrt{200}~=\sqrt{10}\sqrt{2m}~$ (as per diagram)

$\Rightarrow 10\times ~1.414$

$\therefore Displacement=~14.14m$

3. A train starting from a railway station and moving with uniform acceleration attains a speed$40km{{h}^{-1}}$ in $10\min $. Find its acceleration.

Ans: Since the train starts from rest (railway station) = u = zero

Final velocity of train $=v=40km{{h}^{-1}}$

$\Rightarrow \dfrac{(40\times 1000)}{60\times 60m{{s}^{-1}}}=100/9m{{s}^{-1}}$

$\Rightarrow 11.11m{{s}^{-1}}$

\[\begin{array}{*{35}{l}} time\left( t \right)=10min=10\times 60=600\text{ }seconds \\ ~ \\ \end{array}\]

\[Since\text{ }a\text{ }=\dfrac{\left( v~u \right)}{t}~=11.11m{{s}^{-1}}/600\sec =0.018m/{{s}^{2}}~\]

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

Ans: If the objectโs distance time graph is a straight line parallel to the time axis indicates that with increasing time the distance of that object is not increasing hence the object is at rest that is not moving.

Distance-time graph of an object at rest

5. 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: Such a graph indicates that the object is travelling with uniform velocity.

Speed time graph of an object moving with uniform speed

6. 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: $u=90km{{h}^{-1}}=\dfrac{(90\times 1000)}{60\times 60}=25m{{s}^{-1}}$

\[Given:a=-0.5m{{s}^{-2}},v=0\left( \mathbf{trainis}~\mathbf{brought}~\mathbf{to}~\mathbf{rest} \right)\]

\[v=u+at=25+(-0.5)\times t\]

\[\Rightarrow 0=25-0.5x\]

\[\Rightarrow 0.5t=25,ort=\dfrac{25}{0.5}=50\sec \]

\[s=ut+\dfrac{1}{2}a{{t}^{2}}=25\times 50+\dfrac{1}{2}\times (-0.5)\times {{50}^{2}}\]

\[\Rightarrow 1250-625=625m\]

$\therefore s=625m$

7. A stone is thrown in a vertically upward direction with a velocity of$5m{{s}^{-1}}$ . If the acceleration of the stone during its motion is$10m{{s}^{-1}}$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Ans: Given:$u=5m{{s}^{-1}},a=-10m{{s}^{-2}}$

$v=0~~~\left( \mathbf{sinceat}\text{ }\mathbf{maximum}\text{ }\mathbf{height}\text{ }\mathbf{its}\text{ }\mathbf{velocity}\text{ }\mathbf{will}\text{ }\mathbf{be}\text{ }\mathbf{zero} \right)$

$v=u+at=5+(-10)\times t$

$\Rightarrow 0=5-10t$ $10t=5,or,t=\dfrac{5}{10}=0.5\sec $

$S=ut+\dfrac{1}{2}a{{t}^{2}}=5\times 0.5+\dfrac{1}{2}\times (-10)\times {{0.5}^{2}}$

$\Rightarrow 2.5-1.25=1.25m$

$\therefore S=1.25m$

8. Derive the second equation of motion$S=ut+\dfrac{1}{2}a{{t}^{2}}$ graphically?

Ans: Let at time \[\mathbf{T}=\mathbf{0}\] body moves with initial velocity u and at time โtโ body has final velocity โvโ and at time โtโ it covers a distanceโs.

\[\mathbf{AC}=\mathbf{v},\text{ }\mathbf{AB}=\mathbf{u},\text{ }\mathbf{OA}=\mathbf{t},\text{ }\mathbf{DB}=\mathbf{OA}=\mathbf{t},\text{ }\mathbf{BC}=\mathbf{AC}-\mathbf{AB}=\mathbf{V}-\mathbf{u}\]

Area under a \[v-t\]curve gives displacement so,

\[\mathbf{S}=\mathbf{Area}\text{ }\mathbf{of}\text{ }\Delta \mathbf{DBC}+\mathbf{Area}~\mathbf{of}\text{ }\mathbf{rectangle}\text{ }\mathbf{OABD}\to \left( \mathbf{i} \right)\]

\[Area\text{ }of~\Delta DBC=\dfrac{1}{2}\times Base\times Height\Rightarrow \dfrac{1}{2}\times ~DB\times BC\]

\[\Rightarrow \dfrac{1}{2}\times t\times (v-u)\to (ii)\]

\[Area\text{ }of\text{ }rectangle\text{ }OABD\text{ }=\text{ }length\times Breadth\]

\[\Rightarrow OA\times BA\]

\[\Rightarrow t\times u\to (iii)\]

\[S=ut+\dfrac{1}{2}\times t\times (v-u)\]

\[S=ut+\dfrac{1}{2}\times t\times at(\therefore useV-u=at)\]

\[\therefore S=ut+\dfrac{1}{2}a{{t}^{2}}\]

9. A car moving with a certain velocity comes to a halt if the retardation was$5m/{{s}^{2}}$ find the initial velocity of the car?

Ans: \[\mathbf{V}=\mathbf{0}\left( \mathbf{comes}\text{ }\mathbf{to}\text{ }\mathbf{rest} \right)\mathbf{V}=\mathbf{final}\text{ }\mathbf{velocity}\]

$S=62.5m$

$a=-5m/{{s}^{2}}(retardation)$

$U=?$

From \[{{3}^{rd}}\] equation of motion,

${{v}^{2}}-{{u}^{2}}=2as$

${{O}^{2}}-{{u}^{2}}=2\times (-5)\times 62.5$

$-{{u}^{2}}=-10\times 62.5$

${{u}^{2}}=625$

$u=\sqrt{625}[u=25m/s]$

10. Two cars A and B are moving along in a straight line. Car A is moving at a speed of $80kmph$ while car B is moving at a speed of $50kmph$in the same direction, find the magnitude and direction of

Five v the relative of car A with respect to B.

The relative velocity of car B with respect to A.

Ans: Velocity of car A = $80kmph$

Velocity of Car B = \[-\text{ }50~~kmph\]

(-ve sign indicates that Car B is moving in the opposite direction to Car A )

The relative velocity of car A with respect to B

velocity of car A + (- velocity of car B)

$\Rightarrow 80+(-(-50))$

$\Rightarrow 80+50$

$\Rightarrow +130kmph$

$+130kmph$shows that for a person in car B, car A will appear to move in the same direction with a speed of the sum of their individual speed.

Relative velocity of car B with respect to A

Ans:

$\Rightarrow $ Velocity of car B+ (- velocity of car A)

\[ \Rightarrow -50~~+~\left( -80 \right) \]

\[ \Rightarrow -130kmph \]

It shows that car B will appear to move with \[130~~~kmph\] in opposite direction to car A

11. A ball starts from rest and rolls down $16m$down an inclined plane in$4s$

What is the acceleration of the ball?

Ans: Given: \[\mathbf{u}=\text{ }\mathbf{initial}\text{ }\mathbf{velocity}\text{ }=\text{ }\mathbf{0}\] (body starts from rest)\[S=\text{ }distance\text{ }=\text{ }16\text{ }m\text{ }\]

\[T=\text{ }time\text{ }=\text{ }4s\]

From, $s=ut+\dfrac{1}{2}a{{t}^{2}}$

$16=0\times t+\dfrac{1}{2}a\times {{\left( 4 \right)}^{2}}$

$16=\dfrac{1}{2}\times a\times 16$

$\dfrac{16\times 2}{16}=a$

$[2m/{{s}^{2}}=a]$

What is the velocity of the ball at the bottom of the incline?

Ans: From, $v=u+at$

$v=0+2\times 4$

$[v=8m/s]$

12. Two boys A and B, travel along the same path. The displacement โ time graph for their journey is given in the following figure.

How far down the road has B travelled when A starts the journey?

Ans: When A starts his journey at\[4\text{ }sec\] , B has already covered a distance of \[857m\].

Without calculation, the speed, state who is traveling faster A or B?

Ans: A travels faster than B because A starts his journey late but crosses B and covers more distance then B in the same time as B

What is the speed of A?

Ans:

\[Speed\text{ }of\text{ }A\text{ }=\text{ }\dfrac{Distance\text{ }covered}{Time\text{ }taken}\]

Let at \[t\text{ }=12\text{ }min\],\[distance\text{ }covered=3500m\]

$\Rightarrow \dfrac{3500}{12}=375m/\min $

$\therefore Speed\text{ }of\text{ }A=375m/\min $

What is the speed of B?

Ans:

\[Speed\text{ }of\text{ B}=\dfrac{Distance\text{ }covered}{Time\text{ }taken}\]

\[{{V}_{B}}=\dfrac{3000}{12}\]

\[{{V}_{B}}=214m/\min \]

Are the speed of A and B uniform?

Ans: Yes

What does point X on the graph represent?

Ans: X on the graph represents the point at which both A and B are at the same position

What is the speed of approach of A towards B? What is the speed of separation of A from B?

Ans: Speed of approach of A towards \[B=375\text{ }m/min-214\text{ }m/min\]

\[\Rightarrow 161\text{ }m/min\]

Speed of separation of A from \[B=161\text{ }m/min\]

13. A body is dropped from a height of$320m$. The acceleration due to the gravity is $10m/{{s}^{2}}?$

How long does it take to reach the ground?

Ans: Given Data: Height = h

\[Distance=s=320m\]

Acceleration due to gravity $=g=10m/{{s}^{2}}$

Initial velocity \[=u=0\]

From $s=ut+\dfrac{1}{2}a{{t}^{2}}$

$h=ut\times \dfrac{1}{2}g{{t}^{2}}$

$320=0\times t+\dfrac{1}{2}\times 10\times {{t}^{2}}$

$\dfrac{320\times 2}{10}={{t}^{2}}$

$64={{t}^{2}}$

$t=8\sec $

What is the velocity with which it will strike the ground?

Ans: From $v=u+at$

$v=0+10\times 8$

$v=80m/s$

14. Derive third equation of motion${{v}^{2}}-{{u}^{2}}=2as$numerically?

Ans: We know,

$v=u+at......(i)$

$s=ut+\dfrac{1}{2}a{{t}^{2}}......(ii)$

When, v= final velocity

u= initial velocity

a = acceleration

t = time

s = distance

From equation (i) $t=\dfrac{v-u}{a}$

Put the value of t in equation (ii)

$s=u\times \dfrac{v-u}{a}+\dfrac{1}{2}a\times \dfrac{v-u}{a}$

$s=\dfrac{uv-{{u}^{2}}}{a}+\dfrac{1}{2}a\times \dfrac{{{v}^{2}}+{{u}^{2}}-2uv}{{{a}^{2}}}$

$s=\dfrac{uv-{{v}^{2}}}{a}+\dfrac{1}{2}\times \dfrac{{{v}^{2}}+{{u}^{2}}-2uv}{a}$

$s=\dfrac{2uv-2{{u}^{2}}+{{v}^{2}}+{{u}^{2}}-2vu}{2a}$

$s=\dfrac{{{v}^{2}}-{{u}^{2}}}{2a}$

$2as={{v}^{2}}-{{u}^{2}}$

${{v}^{2}}={{u}^{2}}+2as$

16. The velocity-time graph of the runner is given in the graph.

What is the total distance covered by the runner in$16s$?

Ans: We know that area under v-t graph gives displacement:

So, Area = distance = s = area of triangle + area of rectangle \[Area\text{ }of\text{ }triangle=\dfrac{1}{2}\times base\times height\]

\[\Rightarrow \dfrac{1}{2}\times 6\times 10\] \[\Rightarrow 30m\]

\[\therefore Area\text{ }of\text{ }triangle=30m\]

\[Area\text{ }of\text{ }rectangle=length\times breadth\]

\[\Rightarrow \left( 16-6 \right)\times 10\]

\[\Rightarrow 10\times 10\]

\[\Rightarrow 100m\]

Total area = \[180m\]

Total distance = \[180m\]

What is the acceleration of the runner at $t=11s$?

Ans: Since at \[t=11sec\], particles travel with uniform velocity so, there is no change in velocity hence acceleration = zero.

17. A boy throws a stone upward with a velocity of $60m/s$ .

How long will it take to reach the maximum height$(g=-10m/{{s}^{2}})$ ?

Ans:\[u=60\text{ }m/s;\text{ g= -10m/}{{\text{s}}^{2}};v=0\]

The time to reach maximum height is

\[\text{v}=\text{u}+\text{at}=\text{u}+\text{gt}\]

\[0=60-10\text{t}\]

\[\text{t}=\dfrac{60}{10}=6\text{s}\]

What is the maximum height reached by the ball?

Ans: The maximum height is:

\[\text{v}^2=\text{u}^2+2\text{gs}\]

\[\text{s}=-\dfrac{\text{u}^2}{2\text{g}}=\dfrac{60^2}{2\times 10}\]

\[\Rightarrow 180 \text{m}\]

How long will it take to reach the ground?

Ans: The time to reach the top is equal to the time taken to reach back to the ground. Thus, the time to reach the ground after reaching the top is\[6s\]or the time to reach the ground after throwing is\[~6+6=12s\].

18. The displacement x of a particle in meters along the x- axis with time โtโ in seconds according to the equation- $X=2m+\left( \dfrac{12m}{s} \right)t$

Draw a graph if x versus t for $t=0$ and $t=5\sec $

Ans: \[X=20m+\left( 12 \right)\text{ }t\]

At \[t=0\text{ }\]

\[X=20+120=12\text{ }m\]

At \[t=1\]

\[X=20+12=32m\]

At \[t\text{ }=2\]

\[X=20+24=44m\]

At \[t=5\]

\[X=20+125=72m\]

What is the displacement comes out of the particles initially?

Ans: At \[T=0\] (initially)

Displacement \[=20m.\]

What is the slope of the graph obtained?

Ans:

\[Slope=\dfrac{{{y}_{2}}-y1}{{{x}_{2}}-{{x}_{1}}}=\dfrac{72-44}{5-2}=\dfrac{28}{3}\]

$\Rightarrow 9.3m/s$

19. The velocity of a body in motion is recorded every second as shown

Time(s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

Velocity(m/s) | 62 | 54 | 48 | 42 | 36 | 30 | 24 | 18 | 12 | 6 | 0 |

calculate the โ

Acceleration

Ans: Acceleration =slope of the velocity time graph

$a=\dfrac{{{V}_{2}}-{{V}_{1}}}{{{t}_{2}}-{{t}_{1}}}$

$a=\dfrac{54-24}{1-6}=\dfrac{30}{-5}=-6m/{{s}^{2}}$

Distances travelled and draw the graph.

Ans: Distance $\Rightarrow S=ut+\dfrac{1}{2}a{{t}^{2}}$

$\Rightarrow 60\times 10+\dfrac{1}{2}(-6)\times {{(10)}^{2}}$

$\Rightarrow 600-300=300m$

20. Draw the graph for uniform retardation โ

Position โ time graph

Ans:

Velocity โ time

Ans:

Acceleration- time

Ans:

21. The displacement โ time graph for a body is given. State whether the velocity and acceleration of the body in the region BC, CD, DE and EF are positive, negative or Zero.

Ans:

For AB, the curve is upward stopping i.e. slope is increasing so velocity is positive and remains the same so, V= +ve but a = 0.

For BC, the curve has still has +ve slope so, V = +ve but velocity is decreasing with respect to time so, a=negative

For CD, both velocity and acceleration are Zero because slope is Zero.

For DE, velocity is the (v is increasing with respect to time) and so is the acceleration is +ve. (v) For EF, velocity is +ve (positive slope of x-t graph) but acceleration is Zero because velocity remains some with time.

AB | BC | CD | DE | EF | |

V | +ve | +ve | 0 | +ve | +ve |

A | 0 | -ve | 0 | +ve | 0 |

22. Derive the third equation of motion$-{{v}^{2}}-{{u}^{2}}=2as$ as graphically?

Ans: Let at time \[t=0\], the body moves with initial velocity u and time at โtโ has final velocity โvโ and in time โtโ covers a distance โsโ

Area under v-t graph gives displacement

\[\mathbf{S}=\mathbf{Area}\text{ }\mathbf{of}\text{ }\Delta \mathbf{DBC}+\mathbf{Area}~\mathbf{of}\text{ }\mathbf{rectangle}\text{ }\mathbf{OAB}D\]

$S=\dfrac{1}{2}\times base\times height+length\times breadth$

$S=\dfrac{1}{2}\times DB\times BC+OA\times AB$

$S=\dfrac{1}{2}\times t\times \left( v-u \right)+t\times u\to (i)$

$Now,v-u=at$

$\dfrac{v-u}{a}=t$

Put the value of โtโ in equation (i)

$s=\dfrac{1}{2}\times (v-u)\times \dfrac{v-u}{a}+u\times \left( \dfrac{v-u}{a} \right)$

$s=\dfrac{uv-{{u}^{2}}}{a}+\dfrac{1}{2}a\times \dfrac{{{v}^{2}}+{{u}^{2}}-2uv}{{{a}^{2}}}$

$s=\dfrac{{{(v-u)}^{2}}2u(v-u)}{2a}$

$s=\dfrac{{{v}^{2}}+{{u}^{2}}-2uv+2vu-2{{u}^{2}}}{2a}$

$s=\dfrac{{{v}^{2}}-{{u}^{2}}}{2a}$

$2as={{v}^{2}}-{{u}^{2}}$

Third equation of motion

Long Answer Questions (5 Marks)

1. 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 $2\min 20\sec ?$

Ans: circumference of circular track \[=2\pi r\]

$\Rightarrow 2\times \dfrac{22}{7}\times \dfrac{diameter}{2}$

$\Rightarrow 2\times \dfrac{22}{7}\times \dfrac{200}{2}=\dfrac{4400}{7}m$

Rounds completed by athlete in \[2min20sec=s=\dfrac{140}{40}=\text{ }3.5\]

Therefore, total distance covered $=\dfrac{4400}{7}\times 3.5=2200m$

since one complete round of circular track needs \[40s\] so he will complete \[3rounds\text{ }in\text{ }2mins\]and in next \[20s\] he can complete half round Therefore, displacement \[=diameter=200m\].

2. Joseph jogs from one end A to the other end B of a straight $300m$road in $2\min 50\sec $ and then turns around and jogs $100m$back to point C in another$1\min $ . What are Josephโs average speeds and velocities in jogging

From A to B

Ans: \[Distance=300m\]

\[time=2min30seconds=150\text{ }seconds\]

Average speed from A to B = average velocity from A to B

\[\Rightarrow 300m/150s=2m/s\]

From A to C?

Ans: average speed from A to C \[=\dfrac{\left( 300+100 \right)\text{ }m}{\left( 150+60 \right)\text{ }sec}\]

$\Rightarrow \dfrac{400m}{210s}=1.90m/s$

Displacement from A to C \[=\left( 300100 \right)m=200m\]

\[time=2min30sec+1min\text{ }=210s\]

\[velocity\text{ }=\dfrac{\text{ }displacement}{time}=\dfrac{200m}{210s}=0.95m/s\]

3. 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 Abdulโs trip?

Ans. If we suppose that distance from Abdulโs home to school = x km s

while driving to school:$speed=20km{{h}^{-1}}$ ,

\[velocity\text{ }=\text{ }\dfrac{displacement}{time}\]

\[20\text{ }=\dfrac{x}{t},or,\text{ }t=\dfrac{x}{20}hr\]

on his return trip: \[speed=40km{{h}^{-1}},40=\text{ }x/t\]

or, \[t=\dfrac{x}{40}hr\]

\[total\text{ }distance\text{ }travelled=\text{ }x+x=2x\]

\[total\text{ }time\text{ }=\text{ }t\text{ }+\text{ }t=\dfrac{x}{20}+\dfrac{x}{40}=\dfrac{\left( 2x\text{ }+\text{ }x \right)}{40}=\dfrac{3x}{40}hr\]

average speed for Abdulโs trip \[=\text{ }\dfrac{2x}{\left( \dfrac{3x}{40} \right)\text{ }}=\dfrac{80x}{3x}=26.67km/hr\]

4. 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: since the motorboat starts from rest so \[u=0\]

Given data:

$time(t)=8s,a=3m/{{s}^{2}}$

$distacnce(s)=ut+\dfrac{1}{2}a{{t}^{2}}$

\[distacnce(s)=0+\dfrac{1}{2}\times 3\times {{8}^{2}}\]

$\therefore distacnce(s)=96m$

5. A driver of a car travelling at applies the brakes and accelerates uniformly in the opposite direction. The car stops in$5s$. Another driver going at $3km{{h}^{-1}}$ in another car applies is 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:

As given in the figure below AB (in red line) and CD(in red line) are the Speed-time graph for given two cars with initial speeds \[52km{{h}^{-1}}\] and $3km{{h}^{-1}}$ respectively. Distance Travelled by first car before coming to rest=Area of $\Delta OAB$

\[\Rightarrow \left( \dfrac{1}{2} \right)\times OB\times OA\]

\[\Rightarrow \left( \dfrac{1}{2} \right)\times 5s\times 52km{{h}^{-1}}\]

\[\Rightarrow \left( \dfrac{1}{2} \right)\times 5\times \left( \dfrac{52\times 1000}{3600} \right)m\]

\[\Rightarrow \left( \dfrac{1}{2} \right)\times 5\times \left( \dfrac{130}{9} \right)m\]

\[\Rightarrow \dfrac{325}{9}m\]

\[\Rightarrow 36.11m\]

Distance Travelled by second car before coming to rest=Area of $\Delta OCD$

\[\Rightarrow \dfrac{1}{2}\times OD\times OA\]

\[\Rightarrow \dfrac{1}{2}\times 10s\times 3km{{h}^{-1}}\]

\[\Rightarrow \dfrac{1}{2}\times 10\times \left( \dfrac{3\times 1000}{3600} \right)m\]

\[\Rightarrow \dfrac{1}{2}\times 10\times \left( \dfrac{5}{6} \right)m\]

\[\Rightarrow 5\times \left( \dfrac{5}{6} \right)m\]

\[\Rightarrow \dfrac{25}{6}=4.16m\]

$\therefore $ Clearly the first car will travel farther \[\left( 36.11\text{ }m \right)\] than the first car \[\left( 4.16\text{ }m \right).\]

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

Which of the three is travelling the fastest?

Ans: It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.

Are all three ever at the same point on the road? Fig. 8.11

Ans: All of them never come at the same point at the same time.

How far has C travelled when B passes A?

Ans: According to graph; each small division shows about\[0.57\text{ }km\] .

A is passing B at point S which is in line with point P (on the distance axis) and shows about \[9.14\text{ }km\]

Thus, at this point C travels about

\[\text{9}\text{.14-(0}\text{.57}\times \text{3}\text{.75)km= }9.14\text{ }km\text{ }\text{ }2.1375\text{ }km=7.0025\text{ }km\approx 7km\]

Thus, when A passes B, C travels about\[7\text{ }km\].

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

Ans: B passes C at point Q at the distance axis which is \[\approx 4km+0.57km\times 2.25=\text{ }5.28\text{ }km\] Therefore, B travelled about \[5.28\text{ }km\] when passes to C.

7. 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: Let us assume, the final velocity with which ball will strike the ground be 'v' and time it takes to strike the ground be 't'

Initial Velocity of ball \[u=0\]

Distance or height of fall \[s=20\text{ }m\]

Downward acceleration $a=10m{{s}^{-2}}$

As we know,${{v}^{2}}={{u}^{2}}-2as$

$or,2as={{v}^{2}}-{{u}^{2}}$

${{v}^{2}}=2as+{{u}^{2}}$

$\Rightarrow 2\times 10\times 20+0$

$v=\sqrt{400m{{s}^{-1}}}$

$\therefore $ Final velocity of ball,$v=20m{{s}^{-1}}$

\[t=\text{ }\dfrac{\left( v-u \right)}{a}\]

$\therefore $Time taken by the ball to strike $=\dfrac{\left( 20-0 \right)}{10}$

\[\Rightarrow \dfrac{20}{10}\]

\[\therefore Time\text{ }taken\text{ }by\text{ }the\text{ }ball\text{ }to\text{ }strike=2\sec \]

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

Find how far does the car travel 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 car in the \[4\text{ }second\]

The area under the slope of the speed โ time graph gives the distance travelled by an object. In the given graph

\[56\] full squares and \[12\] half squares come under the area slope for the time of \[4\text{ }second\].

Total number of squares \[=56+\dfrac{12}{2}=62\] squares

The total area of the squares will give the distance travelled by the car in \[4\text{ }second\] . on the time axis,

\[5\text{ }squares\text{ }=2seconds,\text{ }therefore\text{ }1\text{ }square\text{ }=\dfrac{2}{5}seconds\]

on speed axis there are \[3\text{ }squares=2m/s\]

therefore, area of one square $=\dfrac{2}{5}s\times \dfrac{2}{3}m/s=\dfrac{4}{15}m$

so area of \[62\text{ }squares=\dfrac{4}{15}m\times 62=\dfrac{248}{15}m=16.53m\]

Hence the car travels \[16.53m\] in the first\[4\text{ }seconds\].

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

Ans: The straight line part of graph, from point A to point B represents a 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

Ans: An object with a constant acceleration can still have the zero velocity. For example, an object which is at rest on the surface of earth will have zero velocity but still being acted upon by the gravitational force of earth with an acceleration of \[9.81\text{ }m{{s}^{-2}}\]towards the center of earth. Hence when an object starts falling freely can have constant acceleration but with zero velocity.

b) An object moving in a certain direction with acceleration in the perpendicular direction.

Ans: When an athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. Here, the motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion where acceleration is always perpendicular to direction of motion of an object at a given instance. Hence it is possible when an object moves on a circular path.

10. An artificial satellite is moving in a circular orbit of radius \[42250km\] . Calculate its speed if it takes \[24hrs\]to revolve around the earth.

Ans. Let us assume An artificial satellite, which is moving in a circular orbit of radius \[42250\text{ }km\] covers a distance 's' as it revolve around earth with speed 'v' in given time 't' of \[24\text{ }hours\] .

\[\Rightarrow 42250\text{ }km\]

Radius of circular orbit r

$\Rightarrow 4225\times 1000m$ Time taken by artificial satellite \[t=\text{ }24\text{ }hours\]

$\Rightarrow 24\times 60\times 60s$ Distance covered by satellite s=circumference of circular orbit

\[\Rightarrow 2\pi \text{ }r\]

โด Speed of satellite\[v=\dfrac{\left( 2\pi \text{ }r \right)}{t}\]

$\Rightarrow \dfrac{\left[ 2\times \left( \dfrac{22}{7} \right)\times 42250\times 1000 \right]}{\left( 24\times 60\times 60 \right)}$

$\Rightarrow \dfrac{\left( 2\times 22\times 42250\times 1000 \right)}{\left( 7\times 24\times 60\times 60 \right)m{{s}^{-1}}}$

$\Rightarrow 3073.74m{{s}^{-1}}$

$\therefore Speed=3.073km/s$

11. The position of a body at different times are recorded in the table given below:

Draw the displacement time graph for the above data?

Ans:

What is the slope of the graph?

Ans:

\[Slope\text{ }of\text{ }the\text{ }graph=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]

\[\Rightarrow \dfrac{\left( 36-24 \right)m}{\left( 6-4 \right)\sec }=\dfrac{12m}{2\sec }\]

\[\therefore Slope\text{ }of\text{ }the\text{ }graph=6m/sec\]

What is the speed of the motion?

Ans: Slope of the graph of a displacement-time graph = speed

Hence speed \[=6m/sec\].

## Important Questions of Chapter Motion Class 9 โ PDF Download

Science being one of the broader and more stringent subjects requires much practice and preparation. Furthermore, Chapter motion also plays a key role in further academics as its concepts are involved in all the higher classes for science stream students. The students are thus advised to refer to the Class 9th Science Chapter 7 Important questions.

The PDF version of Class 9th Science Ch 7 Important questions is available easily on Vedantu website. The students can access them for free online, and can also download it on any device for future reference. Important questions for Class 9 Science Chapter Motion helps the students get the best practice and preparation for the examinations, and pass the exams with flying colours.

Vedantu has a team of experts with immense Science qualifications who design the PDF of Ch 7 Motion Class 9 important questions, keeping the past statistics and records in mind. The solutions of the problems are also available along with them, to provide the best assistance to the students. During the examination times, and when the students need the perfect resources for practice during revision, they must refer to these essential questions and get well-versed in the type of questions asked in the examinations, and the essential concepts related to them.

### Chapter 7 Class 9 Science Important Questions โ Related Essential Concepts

Important questions for Class 9 Science Chapter 7 are specifically designed using Chapter Motion's essential topics and sub-topics. Here is a brief on all the vital concepts and related topics involved in the chapter.

### Motion

The motion refers to the movement of anybody from one position to another concerning the object's observer. When the object's motion goes in a straight line, it is known as a rectilinear motion or motion in a straight line. For example, when a car moves on a straight highway for some distance, it is known as a rectilinear motion.

### Motion is of Two Types

Uniform Motion โ Uniform motion refers to the one in which the object covers equal distance in an equal time interval.

Non-Uniform Motion โ Non-uniform motion refers to the one in which the object covers the unequal distance in an equal time interval.

### Scalar and Vector Quantities

A scalar quantity is the one whose direction is not associated with it. For example, time, mass, etc. On the other hand, a vector quantity is the one that has both direction and magnitude associated with itโfor example, position, force, etc.

### Distance and Displacement

Distance is the length of the path that any moving object covers in the given period, and it is independent of its direction. Distance is a scalar quantity, and its SI unit is metre (m).

In contrast, Displacement is the shortest distance between the initial and final position of the object. The unit of Displacement is same as that of distance, but it is a vector quantity.

### Speed and Velocity

Speed refers to the distance per unit time for any object. Speed is a scalar quantity, and its SI unit is metre/ second. For any given non-uniform motion under consideration, the average speed for any object is the total distance covered in the total time taken.

On the other hand, Velocity is a vector quantity, and it refers to the speed of any object travelling in a particular direction. The Velocity of any object is the Displacement per unit time. Average Velocity is the arithmetic mean of the final and initial Velocity for a certain period.

### Acceleration

Acceleration is the rate of change of Velocity. Acceleration for an object is taken as the difference between the final Velocity and initial Velocity per unit time. SI unit of acceleration is metre / second2. Acceleration is a vector quantity, and it is positive when the acceleration is in the direction of Velocity, else vice versa. Retardation or deceleration is the term used to describe the negative acceleration.

### Uniform and Non-Uniform Acceleration

When there is a change of Velocity for an object by an equal amount for an equal time period, the acceleration is uniform. For example, the motion of a freely falling object is in uniform acceleration.

On the other hand, non-uniform acceleration is when the object's velocity changes unequally in equal time intervals. For example, the motion of a vehicle on the road is in non-uniform acceleration.

### More Topics Related to Important Questions for Class 9 Science Chapter 7

Class 9 Science Chapter 7 extra questions also carry some weightage from the other topics like the derivation of different motion laws and graphical representation of motion. And the students going through the entire set of questions get the best practice and score well in the examinations.

### Motion Class 9th Important Questions

The expert team from Vedantu thoroughly researched the previous year question papers and statistics. It came up with the set of **Class 9 Science Chapter 7 important questions** to enhance the students' practice and preparation for school and competitive examinations. Here are the ten important questions of Chapter Motion Class 9:

Derive the three equations of motion using graphical representation and arithmetic expressions.

The Displacement of an object in the given time interval is 0, would the distance travelled also be zero? Justify the answer with examples.

Explain how there will be a change in the motion equations for an object moving with a uniform velocity.

A car starts from rest and moves towards the x-axis with a constant acceleration of 5 m/s for 8 seconds. If the car continues with a constant velocity, calculate the distance it will travel in 12 seconds.

A person moves from a given point A to another point B with a uniform speed of 30 km/h, and it then returns with the speed 20 km/h. Find the average speed of the person.

Draw the velocity-time graph of a piece of item thrown upwards and then it's way back downwards after reaching the maximum height.

Calculate the relation of distance that an object travelled while moving with uniform acceleration in the given time interval between the 4th and the 5th second.

A ball drops from rest at height 150 m and at the same time other drops from rest at height 100m. Calculate the difference in their heights after the end of 2 s if both of them drop with the same accelerations. How does the difference in heights vary with the given period?

A car started from rest travels 30 m in the first 2 seconds and then 170 m in the next 4 seconds. Calculate the Velocity after 7 seconds from the starting.

An electron moves with a velocity of 5 * 104 m/s, and it enters a uniform electric field and acquires uniform acceleration of 104 m/s2, in initial motion's direction. For this situation:

How much distance will be covered by the electron in that time?

Calculate the time by when the electron will acquire the velocity double of its initial Velocity.

### Benefits of Important Questions for Class 9 Science Chapter 7

Motion Class 9 important questions help the students in their preparations, and they are very beneficial for the students, because:

Having the

**Class 9 Science Chapter 7 important questions**handy helps the students get the instant practice material during the revision times before the examinations.Precisely designed Class 9 Science Motion Important questions carry all the questions that either occurred in the

**previous year papers**or have similar problem statements. Thus they help the students get an idea about the upcoming papers.The students practising through the important questions, get a confidence boost for the examinations and prepare for solving all types of problems in the examinations.

### Conclusion

Crucial Class 9 Science Chapter Motion questions are simply available on Vedantu websites and are vital and valuable for students' preparations. After extensive investigation, an expertly created set of critical questions provides students with a compact study resource that serves as the ideal revision material during test timings. It is required for both school and competitive examinations and provides with an understanding of the sorts of all-important and commonly asked questions.

## Important Related Links for CBSE Class 9

CBSE Class 9 Study Materials |

Important Questions for CBSE Class 9 |

## FAQs on Important Questions for CBSE Class 9 Science Chapter 7 - Motion

1. What are some important questions of Chapter 7 of Class 9 Physics?

All of the relevant questions for Chapter 7 of Class 9 Physics may be found on the Vedantu website and mobile app. Pupils may also obtain free PDFs of crucial problems for all chapters in Science for Class 9. This chapter contains a variety of significant test subjects, so students will benefit from reading and answering all of the important questions when studying for the exam and revising.

Some of the important questions are;

What is velocity?

Define Speedย

What is odometerย

What is uniform acceleration?

2. If an object has travelled a distance can its displacement be zero?

Displacement is determined by a change in the position of a body from point A to point B. In the case where point B is the same as Point A, the displacement would be zero. For example, if a boy moves from his home to go to the market and comes back home after covering a distance of 2 km, the displacement would be zero. This is because the initial starting point and the point after the movement are the same.

3. What is motion (long answer)?

Motion is defined as a change in a body's or object's location with relation to its surroundings and time. The two forms of motion are uniform and non-uniform motion. This category includes speed, velocity, distance, displacement, and acceleration. The revision notes and crucial questions for this chapter are available in offline PDF format from the Vedantu website and mobile app. The answers to the main questions are selected in a way that follows the intended reply strategy.

4. How do you solve numericals in Chapter 7 of Class 9 Science?

To solve the numerical problems in this chapter, you must first memorise all of the formulae and subjects presented in the chapter. A thorough knowledge of the ideas and formula derivations leads to a better understanding of the numerical difficulties. Students can use the PDF for Key Questions from Vedantu to obtain some assistance on how to apply each formula to different sorts of numericals. The PDF includes one of each type of question, ranked by significance.

5. How does solving the important questions help for Chapter 7 of Class 9 Physics?

The best strategy to revise the chapter is to answer the essential questions. Given the lengthy course, it is crucial to ensure that no important topic is overlooked for preparation. These questions provide you the opportunity to go over all of the chapter's themes in a short amount of time. It saves you time and energy because instead of going through all of the questions, you simply look at the ones that are important.