Important Questions for CBSE Class 9 Science Chapter 8 – Motion
CBSE Class 9 Science Chapter 8 Motion Important Questions
Science is a foundational subject for students. Students must thoroughly understand the subject to a level required for both school and competitive exams.
Chapter 8 of Class 9 Science is about motion. Motion is a part of our routine. Running, walking, and flying is all motion. Galaxy, planets, and stars are all in motion.
Extramarks is a leading online learning platform trusted by many students and teachers across the country. Students can visit the Extramarks website to access our comprehensive suite of study materials, including NCERT solutions, CBSE sample papers, revision notes, etc. Our team of Science expert teachers have created these study notes by referring to the latest CBSE curriculum.
Class 9 Science starts becoming challenging, and the complexity of Physics, Chemistry and Biology requires proper understanding of concepts and continuous revision. At Extramarks, our team understands the importance of regularly solving examoriented questions. Students can refer to our question bank Important Questions Class 9 Science Chapter 8, which consists of questions of various levels of difficulty from sources such as NCERT textbooks, exemplar books, past year question papers, reference books, etc.
Solving Important Questions Class 9 Science Chapter 8 allows students to understand whether they have covered the topic altogether or not. If students have missed something, they can learn it while solving the questions from our Class 9 Science Chapter 8 Important Questions study material
Get Access to CBSE Class 9 Science Important Questions 202223 with Solutions.
Also, get access to CBSE Class 9 Science Important Questions for other chapters too:
CBSE Class 9 Science Important Questions  
Sr No  Chapters  Chapter Name 
1  Chapter 1  Matter in Our Surroundings 
2  Chapter 2  Is Matter Around Us Pure 
3  Chapter 3  Atoms and Molecules 
4  Chapter 4  Structure of Atom 
5  Chapter 5  The Fundamental Unit of Life 
6  Chapter 6  Tissues 
7  Chapter 7  Diversity in Living Organisms 
8  Chapter 8  Motion 
9  Chapter 9  Force and Laws of Motion 
10  Chapter 10  Gravitation 
11  Chapter 11  Work and Energy 
12  Chapter 12  Sound 
13  Chapter 13  Why Do We Fall ill 
14  Chapter 14  Natural Resources 
15  Chapter 15  Improvement in Food Resources 
CBSE Class 9 Science Chapter 8 Question Answers
Solving the Extramarks Important Questions Class 9 Science Chapter 8 question paper is important for students to gain a good hold on the chapter. Our question bank contains various questions, including long questions, short questions, MCQs, and fillintheblanks, all based on the latest CBSE syllabus. Students will come to know how they should present an answer to score maximum marks.
Students are advised to regularly solve questions from our Important Questions Class 9 Science Chapter 8 as often as possible. It helps understand the concept and learn complex topics in a straightforward manner and with thorough revision.
Question 1. What are the differences between scalar and vector quantities? Give examples.
Answer 1:
Scalar quantities  Vector quantities 


For example, a car travels at a speed of 20km/h 
For example, increase/ decrease in temperature 








Time and distance Speed Density Energy Mass Length 
Displacement Velocity Acceleration Weight Force 
Question 2. Differentiate between uniform and nonuniform motion. Give examples.
Answer 2:
Uniform motion  Non–uniform motion 
Type of motion in which an object covers an equal distance in an equal interval of time  Type of motion is when an object covers an unequal distance in an equal time interval 
It is similar to the speed of the object.  It is different from the speed of the object. 
Has zero acceleration.  Has nonzero acceleration. 
Example:

Example:

The distancetime graph shows a straight line inclined on the xaxis.  The distancetime graph shows a curved line. It can also be a zigzag. 
Answer 3: Graphs are important because they:
 visually represent the data.
 present information quickly.
 For example, the velocitytime graph gives us information about the velocity of an object at any time instance, total distance travelled, total time and acceleration of the body.
Question 4. Represent a distancetime graph for uniform and nonuniform motion of an object.
Answer 4: In the distancetime graph, time is taken on the xaxis and distance is taken on the yaxis. The graph depicts the distance covered by the object during a particular time interval.
In uniform motion, the uniform speed of the object moving is equal to the slope of the plot, which is a straight line. The graph has a varying slope when the object is in nonuniform motion.
When the distance decreases with time in nonuniform motion, the slope is negative.
Question 5. An object has moved through a distance. Can it have zero displacements? Support your answer with an example.
Answer 5: Yes, the displacement can be zero. Displacement is defined as the distance travelled between the initial and final positions. The displacement is zero if the object returns to the initial point after travelling a distance. To elaborate further,
Average velocity = Net displacement/time taken,
Average speed = Total distance/time taken.
Net displacement can be zero, but total distance cannot be zero. This also implies that average velocity can be zero, but average speed cannot be zero.
Question 6. How are the states of rest and motion relative?
Answer 6: Rest and motion are relative terms. The state of motion and rest depend upon the observer’s frame of reference. They are interpreted based on the change in position of the object concerning the origin. For example, a person standing on the platform will see the train’s passengers in a state of motion. It means we see motion and rest based on the observer’s frame of reference.
Question 7. Why do we put time on the xaxis and distance on the yaxis in the distancetime graph?
Answer 7: This is because an independent variable like time is plotted on the xaxis, and the quantity whose variation depends on our choice like distance, is plotted on the yaxis.
Question 8. Distinguish between speed and velocity.
Answer 8:
SPEED  VELOCITY 
Speed is the distance an object or body travels in a given period.  Velocity is the displacement travelled by an object or a body in a given period. 
Speed depicts the rate of change in distance.  Velocity depicts the rate of change of displacement. 
Average speed = total distance travelled/total time  Average velocity = total displacement/total time 
Speed has magnitude and no direction.  Velocity has both magnitude and direction. 
SI unit is m/s  SI unit is m/s. 
Speed can be equal to velocity.  An object may have the same speed but different velocities. 
Speed = distance/time  Velocity = displacement/time 
It is a scalar quantity.  It is a vector quantity. 
Speed is always positive.  Velocity can be positive or negative. 
Speed cannot be negative and zero.  Velocity can be zero. 
Question 9: Define and classify motion.
Answer 9: Motion is defined as the change in position of an object continuously concerning origin and time. We perceive an object when its position changes with time. The origin is called a fixed reference point. Motion is measured by speed, velocity and acceleration.
Question 10. A farmer moves along a square field of 10 m in 40 s. What will the farmer’s magnitude of displacement be at the end of 2 minutes 20 seconds from his initial position?
Answer 10:
Distance covered by farmer in 40 s = 4 × 10m = 40m
After 40s, farmer is again going to the initial position so displacement = 0
Speed = distance / time
Speed = 40m/40s = 1m/s
Total time = 2 min 20 sec = 140s
1 round completes in 40 s
Three rounds completed in 120 s
Since 120 m is covered in 2 min
Since the farmer covers one round in 40 s therefore he completes ½ round in 20 s.
Therefore, 20 s will cover another 20m
Total distance in 2 min 20 sec = 120 + 20 = 140m
Displacement = √102 + 102 = 10√ 2 m= 10 × 1.414 = 14.14 m
Question 11. Aryan went from Delhi to Chandigarh on his bike. The bike’s odometer reads 4200 km at the start and 4460 km at the end of the trip. If Aryan took 4 hours and 20 min to complete his journey, find the average speed and velocity in km/h and m/s.
Answer 11:
Distance covered = 4460 km – 4200 km
= 260 km
Time = 4 h 20 min = 4 13 h = 13/3 h
Average speed = total distance travelled/ total time taken
= 26013/3 = 60 km/h
Average speed = 60 × 518 = 16.67 m/s
Question 12. A bike can accelerate from rest to 28 m/s in only 4 s.
(a) What is average acceleration?
(b) What distance does it travel in that time?
Answer 12:
Initial velocity = u = 0
Final velocity v = 28 m/s
Time, t = 4 s
To calculate acceleration,
a = vu/t = 28/4 = 7m/s
To calculate distance,
s = ut + ½ at 2
s = 0 + ½ ×7 × 16
s = 56m
Question 13: A bus travels at a speed of 90 km/h. Brakes are applied to produce a uniform acceleration of 0.5 m/s 2. Please find out how far the train will go before it come to rest.
Answer 13:
Initial speed = u = 90km/h
u = (90 × 1000)/60 × 60 = 90000/3600 = 25m/s
a = 0.5 m/s 2
since the bus comes to rest
therefore velocity, v = 0
v = u + at
0 = 25 + (0.5) × t
0 = 25 – 0.5t
0.5t = 25
t = 25/0.5 = 50seconds
s = ut + ½ at2 = 25×50 + ½ × (0.5) × 502
s = 1250 – 625 = 625m
Question 14. Explain negative acceleration. Identify the part of the graph which has negative acceleration
Answer 14: Negative acceleration is also known as retardation or deacceleration. When the velocity of the body decreases with time, the final velocity becomes less than the initial velocity. This is negative acceleration.
For example: when the bus is moving and brakes are applied, there is a gradual decrease in velocity due to retardation.
In the graph, since velocity is decreasing from B to C., therefore, the BC area of the graph represents negative acceleration.
Question 15. Draw a velocitytime graph for a body that has initial velocity ‘u’ and is moving with uniform acceleration ‘a’. Use it to derive v = u + at; s = ut + ½ at2, and v2 = u2 + 2 as
Answer 15:
Slope of acceleration = a = v – u/t
v = u + at
s = area of ABCD
s = 12 (AB + CD) (AD)
s = 12 (u + v) (t)
s = 12 (u + u + at) (t)
s =12 (2u + at)t
s = ut + ½ at2
s = ½ (u + v) (vu)/a = v2 u2/2a
v =u2 + 2as
Question 16. Answer these questions:
(i) An object moves on a circular path of radius r. Calculate the distance and displacement when it completes a half revolution.
(ii) Give the name of the physical quantity, which corresponds to the velocity change rate. Give its SI unit.
(iii) Why is motion in a circle with constant speed called accelerated motion?
Answer 16:
To calculate distance and displacement
 Distance = ½ of circumference = πr
Therefore,
Displacement = diameter = 2r
(ii) Acceleration corresponds to the rate of change in velocity.
The unit of acceleration is m/s 2
(iii) When a change of direction occurs, the velocity of the object changes.
This is called accelerated motion.
Question 17. Two stones are thrown vertically upwards simultaneously with their initial velocities. Prove that the ratio of heights reached by them would be in the proportion of u12:u22. Assume upward acceleration is g and downward acceleration is +g.
Answer 17:
Velocities = u1 and u2
Heights = h1 and h2
Acceleration = g and g
At the highest points,
v1 = v2 = 0
v2 = u2 + 2as
v12 = u12 – 2gh1
v22 = u22 – 2gh2
0 = u12 – 2gh1
0 = u22 – 2gh2
U12 = 2gh1
U22 = 2gh2
u12/u22 = 2gh1/2gh2
Therefore,
u12 : u22 = h1 : h2
Question 18. Draw a velocity versus time graph of a stone thrown vertically upwards and downwards after attaining the maximum height.
Answer 18: The stone has maximum velocity when thrown vertically upwards. When it reaches the maximum height, the velocity decreases and eventually becomes 0. After getting the maximum height, the stone begins to fall. This is the point where velocity begins to increase. The velocity reaches a peak when the rock hits the ground.
Question 19 A car starts from rest and moves with a uniform acceleration of 0.1 m/s 2 for 2 minutes. Find:
(a) The speed acquired
(b) The distance travelled
Answer 19:
 Initial speed = 0
Acceleration = 0.1 m/s 2
Time = 2 min = 120 sec
V = u + at
V = 0 + 0.1 × 120
Speed acquired = 120 m/s
 Distance = ut + ½ at2
S = 0 × 120 + ½ × 0.1× (120)2
S = 720m
Question 20: What is positive and negative slope?
Answer 20: When the speed of the object increases with time, then the distancetime graph has a positive slope.
When the speed of the object decreases with time, then the distancetime graph has a negative slope.
Question 21. A truck starting from rest moves with a uniform acceleration of 5 m/s2. Find its velocity when it has travelled a distance of 1 km.
Answer 21:
Initial velocity, u = 0
Acceleration, a = 5m/s2
Displacement, s = 1 km = 1000m
According to the third equation,
v2 = u2 + 2 as
v = 2as
v = 2×5×1000
v = 100 m/s
Question 22. A racing car has a uniform acceleration of 4 m s 2. What distance will it cover in 10 s after starting?
Answer 23:
Initially, the car is at rest therefore,
Initial velocity, u = 0
Acceleration = a = 4m/s2
Time taken = t = 10s
Distance covered by the racing car, s = ut + ½ at2
S = 0 + ½ × 4 ×(10)2 = 400/2 = 200m
Question 24. A bus decreased its speed from 80 km/ h to 60 km/h in 5 s. Find the acceleration of the bus.
Answer 24:
Initial speed of the bus = u = 80km/h = 80×518 = 22.22 m/s
Final speed = v = 60km/h = 60 × 518 = 16.66 m/s
Time = 5 s (to decrease the speed)
Acceleration= a= vut = 16.622.25 = 1.112 m/s
The negative sign indicates that the velocity is continuously decreasing.
Question 25. The walls of your classroom are in motion but appear stationary. Explain.
Answer 25: Since the relative position remains constant, the classroom walls appear to be at rest to us. But the same classroom seems to be moving to the person in outer space because of the earth’s rotation.
Question 26. State with an example which of the following situations are possible.
(a) An object with a constant acceleration and zero velocity.
(b) An object moving with acceleration but with uniform speed.
(c) An object moving in a particular direction with acceleration in the perpendicular direction.
Answer 26:
(a) When an object is thrown upwards, at the highest point, it comes to rest. The velocity is zero, but the acceleration due to the earth’s gravitational pull still acts on it.
 b) The speed is constant in a uniform circular motion. There is a change in velocity with direction. The centripetal force will always give acceleration in this case.
 c) When an object is thrown forward, its motion is in the horizontal direction. The acceleration due to the earth’s gravity will act vertically downward.
Question 27. When will you say a body is in
(i) uniform acceleration
and (ii) nonuniform acceleration?
Answer 27:
(i) When the velocity of the body changes in equal amount, in equal intervals and in the same direction. Then this is known as uniform acceleration.
The acceleration of an object remains constant and does not change as a function of time.
 When there is a change in velocity in unequal amounts in equal intervals of time, it is known as nonuniform acceleration.
Question 28. Distance is always ___________.
(a) a short length between the two points
(b) a path covered by a body between the two points
(c) the product of length and time
(d) None
Answer 28:
(b) a path covered by a body between two points
Explanation: The actual length of the path between the two points is the distance. It could be a curved, zigzag or straight line. It is equal to displacement when the distance between the two points is a straight line.
Question 29. The area under a vt graph represents a physical quantity which has the unit _____.
 m2
(b) m
(c) m3
(d) ms1
Answer 29: (b) m
The area under the velocity graph represents the displacement and has SI unit metres (m).
Question 30. When the displacement of an object is proportional to the square of time, then the object moves with ________.
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration
Answer 30: (b) uniform acceleration
Explanation:
Acceleration is measured as m/s 2 and velocity as m/s. Therefore, uniform acceleration is the correct answer.
Question 31. A body is thrown vertically upward with velocity u; the most significant height h to which it will rise is,
 u/g
 u2/2g
(c) u2/g
(d) u/2g
Answer 31: (b)
Explanation:
V2 = u2 + 2as
V = 0
a = g
s = H
0 = u2– 2gH
H = u2/2g
Question 32. The numerical ratio of displacement and distance for a moving object is _________.
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal or less than 1
Answer 32: (d) equal or less than 1
Explanation: Displacement of a moving object is never greater than the distance travelled by it.
Therefore Displacement Distance
This implies that Displacementdistance≤1
Displacement is a vector. It may be positive or negative.
Distance is scalar and can never be negative.
Question 33. Fill in the blanks with the appropriate words:
(a) __________ is the other name of negative acceleration.
(b) The acceleration of an object moving with uniform velocity is __________
(c) We should know __________ and __________ to predict the position of a moving body.
(d) The physical quantity which gives an idea of how slow or fast a body is moving is known as __________.
(e) __________ is used to measure the instantaneous speed of the car.
(f) The motion of a bus on a curved road is an example of ______________.
(g) Motion is ___________in nature.
(h) The motion of water during a tsunami is an example of __________.
(i) The earth revolve around the sun in __________.
(j) __________ is given by the area under the velocitytime graph.
(k) Motions are presented in the form of _________.
Answer 33:
(a) Retardation is the other name for negative acceleration.
(b) The acceleration of an object moving with uniform velocity is zero.
(c) We should know speed and direction to predict the position of a moving body.
(d) The physical quantity that gives an idea of how slow or fast a body moves is known as speed.
(e) Speedometer is used to measure the instantaneous speed of the car.
(f) The motion of a bus on a curved road is an example of nonuniform motion.
(g) Motion is relative.
(h) The motion of water during a tsunami is an example of uncontrolled motion.
(i) The earth revolves around the sun in a uniform circular motion.
(j) Displacement is given by the area under the velocitytime graph.
(k) Motions are presented in the form of graphs.
Question 34. Define the displacement of a particle in linear motion. Does it depend upon the origin?
Answer 34: The smallest distance travelled from the initial to the final position of the particle is known as displacement. The displacement does not depend on the choice of origin.
Question 35. A signal from a spaceship reached the ground station in 5 minutes. Calculate the distance of the spaceship from the ground station.
Answer 35: Speed of light = 3 × 108 ms1
Time is taken by signal = 5 min
1 minute = 60 sec
5 min = 5 × 60 sec = 300 sec
Distance of spaceship from ground station = 3 × 108 × 300 = 9 × 1010 m
Question 36. A cyclist travels a distance of 4 km from P to Q and then moves a distance of 3 km at a right angle to PQ. Find his displacement.
Answer:
PQ = 4 km
QR = 3 km
Displacement, PR = √25 = 5 km
Question 37. The motion of four cars A, B, C and D, is represented below. Which of the cars is travelling:
(a) The fastest
(b) The slowest
Answer 37:
(a) C has the highest slope and shows the fastest speed
(b) B has the lowest pitch and shows the slowest travelling speed
Question 38. Suppose a boy is enjoying a ride on a merrygoround which is moving with a constant speed of 10m/s. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity.
Answer 38: (c) in accelerated motion
Speed is constant, but velocity is not consistent in the merrygoround. The direction keeps on changing. There is acceleration in the motion, so the answer is the boy is in accelerated motion.
Question 39. Observe the signboards on roads indicating the speed limit. What does this indication mean? Why is speeding a hazard?
Answer 39: The speed limit is the distance a vehicle running on the road can safely cover in 1 hour. Overspeeding threatens life and can lead to severe accidents and injuries.
Question 40. A car has a uniform acceleration of 4 cm/s 2. What distance will it cover in 10s after the start?
Answer 40:
u= 0, a = 4m/s 2, t = 10 sec
s = ut + ½ at 2
s = 0 + ½ × 4×102
s = 200 m
Question 41. (a) Identify the type of motion in the following examples:
 A car moving with constant speed turns around a curve.
 An electron orbiting around the nucleus.
 An artificial satellite is moving in a circular orbit with a radius of 36000 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer 41:
 i. uniform motion
 nonuniform motion
 V = 2πr = 2 ×227 ×3600024×60×60 = 2.62 km
Question 42. The magnitude of the average velocity is equal to the average speed in which condition?
Answer 42: This is possible when an object travels along a straight line path, and the distance travelled equals displacement.
Question 43. What is the use of the odometer of an automobile?
Answer 43: The distance covered by an automobile is measured by the odometer.
Question 44. What can you say about the motion of an object whose distancetime graph is a straight line parallel to the time axis?
Answer 44: This means that the body is at rest and stationary.
Question 45. The motion of a bee flying randomly in the air is an example of this type of motion.
Answer 45: Nonuniform motion
Question 46. Name the term given to speed in a particular direction.
Answer 46: Velocity
Question 47. Convert the speed of 6m/s into km/h.
Answer 47: 6 m/s = 6 × 3600/1000 = 21.6 km/h
Question 48. Identify the motion in the given graph.
Answer 48:
The graph has a negative slope and depicts uniform retardation.
Question 49. Sohan travels at 20m/s from home to tuition and returns at 25m/s. Find his average velocity.
Answer 49: Since displacement is zero, the average velocity is zero.
Question 50. What causes the phenomena of sunrise, sunset and change of seasons? How do we perceive this cause?
Answer 50: We observe the change in positions of stars, moon, and planets in outer space due to the earth’s motion. Change of season occurs due to the earth’s motion around the sun.
Question 51. What is motion in a straight line?
Answer 51. Motion in a straight line means the body moves along xcoordinate, a straight line. It is also called the onedimensional motion or rectilinear motion of a particle.
Question 52. An artificial satellite is moving in a circular orbit radius of 42250 km. Calculate speed if it takes 24 hours to revolve around the earth.
Answer 52: Distance covered in a revolution = 2r = 2 × 22/7× 42250 km = 265571.43 km
Time of 1 revolution = 24 hour
Speed of satellite = 265571.43/24 = 11065.48 km/h
Question 53. Describe acceleration with examples.
Answer 53. Acceleration is defined as the rate of change of velocity of a moving body with respect to time.
Acceleration = change in velocity/change in time.
It is a vector quantity. Its SI unit is m/s2.
When acceleration is in the direction of the velocity, it is positive. It is negative when opposite to the direction of velocity. For a body with constant velocity, acceleration is zero.
There are two types of acceleration: uniform and nonuniform acceleration.
In uniform acceleration, an object travels in a straight line and the velocity increases or decreases at an equal interval in equal amounts.
When the velocity increases or decreases in an unequal amount in an equal interval of time, then the object is said to be in nonuniform acceleration. It is also known as variable acceleration.
Question 54. Why can a velocitytime graph never be a straight line parallel to the velocity axis?
Answer 54. Velocity changes within a certain period of time. The change is not possible at one time. This would mean that velocity is increasing without an increase in time which shows that acceleration is infinite. This is not possible.
Question 55. Is it possible that the train you are sitting on will appear to move while at rest?
Answer 55: Yes. We are at rest, and the train on the adjacent track is moving.
Question 56. What is the numerical ratio of average velocity to an average speed of an object moving along a straight line path?
Answer 56: The ratio is 1:1 because average velocity equals the average speed in a straight line motion.
Question 57. A car accelerates nonuniformly over a path for time t. Do equations of motion hold in this case? Why/Why not?
Answer 57: No. This is because the car is in nonuniform motion and the equations of motion are applicable for uniform accelerated motion only.
Physics is a subject which requires a proper understanding of terminologies, formula applications and graphs. Students must revise these entities before the exam to ensure they apply formulas and concepts perfectly.
Question 58. Describe equations of motion.
Answer 58.
 Equation for velocity – Time relation or 1st equation of motion
V = u + at
 Equation for position – Time relation or 2nd equation of motion
s = ut + 12 at2
 Equation for position – Velocity relation or 3rd equation of motion
v2 = u2 + 2as
Question 59. Describe graphical presentation of motion.
Answer 59:
 The displacementtime graph
Slope of the displacementtime graph gives the velocity of the object.
Time interval = t2 – t1
Displacement = s2 – s1
Slope = Y interceptX intercept = s2s1t1t1 = velocity
 Velocitytime graph
The graph shows how the velocity changes with time. The slope of the velocitytime graph gives the acceleration of the moving body
Slope of vt graph = Y interceptX intercept = OBOAt2t1 = vut = acceleration
Q60. Describe uniform circular motion.
Answer 60. A body is in a uniform circular motion when it moves with constant speed, following a circular path. The velocity of a particle changes continuously in this type of motion. It is also known as accelerated motion.
Examples:
 The earth’s gravitational force keeps the satellite in orbit.
 The movement of electrons around the nucleus.
 The movement of blades of the windmills.
 The second tip of the watch is in a circular motion.
Question 61. Account for the following:
 Name the quantity measured by the area occupied below the velocitytime graph.
 An object is moving in a certain direction with an acceleration in perpendicular directions.
 Under what condition is the magnitude of the average velocity of an object equal to its average speed?
 An example of uniformly accelerated motion.
 A body is moving along a circular path of radius R. What will the distance and displacement of the body be when it completes a half revolution?
Answer 61. (a) Distance is measured by the area occupied below the velocitytime graph
(b) The motion of the satellite has acceleration perpendicular to the directions.
( c) When distance and displacement are equal, the magnitude of average velocity equals the object’s average speed.
(d) The motion of a freely falling body is an example of uniformly accelerated motion.
(e) The distance is half the circumference. Therefore,
Distance = 2πR2= πR
Displacement = 2R
Question 62. Why is a circulating fan an example of non–uniform motion?
Answer 62. The direction of motion of the circulating fan changes at every point. Therefore, this is non–uniform motion.
Question 63. The velocitytime graph of a body is shown in the figure. Answer the following:
 State the kind of motion represented by 0A and AB.
 What is the body’s velocity after 10 s and after 40 s?
 Calculate the negative acceleration of the body.
 Calculate the distance covered by the body between the 10th and 30th seconds.
Image Source: Oswal CBSE CCE question bank
Answer 63:
 OA represents uniform acceleration, and AB represents zero acceleration or constant velocity.
 The velocity of the body after 10 s is 20 m/s. After 40 s, the body comes to rest.
 Negative acceleration or retardation is calculated as:
Retardation = 0204030 = 2 m/s2
 To calculate the distance between 10th and 30th second
= (30 10)s × 20 m/s = 400m
Question 64. A 100 m long train crosses a 500 m long bridge at 30 m/s. Find the time taken by the bridge to cross it.
Answer 64. Length of train = 100 m
Length of bridge = 500 m
Total length of path covered by train = 500m + 100m = 600 m
Since speed of train is 30 m/s
Therefore, time taken to cover the bridge = distance/speed = 600/30 = 20 s
Question 65. Account for the following:
 What is the shape of the path of a body when it is in uniform motion?
 Give an example of nonuniform motion.
 As shown in the figure, two cars, A And B, have their xt graphs. Which has greater velocity?
Image Source: oswal cbse cce question bank
 A body is moving with a velocity of 15 m/s. If the motion is uniform, what will be the velocity after 15 s ?
Answer 65. (a) The path is a straight line in uniform motion.
(b) A bus moving on a hilly road is an example of nonuniform motion.
(c) Since A has a steeper slope, the slope of velocity is more.
(d) Since the motion is uniform, the velocity is 15m/s even after 15 s.
Question 65. A stone is thrown vertically upwards with an initial velocity of 40 m/s. Find the maximum height reached by the stone. What are the net displacement and the total distance covered by the stone?
Answer 65. Given that
Initial velocity = 40 m/s
Final velocity = 0
v2 = u2 + 2gh
v2 – u2 = 2gh
v2 – u2/2g = h
h = (0)2 – (40)2/ 2 × 10 = +40 × 40/+2 × 10 = 80m
As directions are opposite and the stone returns back to origin, displacement is zero.
Total distance covered = 80 + 80 = 160 m
Question 66. A cheetah can accelerate from rest at the rate of 4m/s2
 What will be the velocity attained by it in 10s?
 How far will it travel in this duration?
Answer 66.
 To calculate velocity attained in 10 s,
Initial velocity = u = 0
Acceleration = a = 4m/s 2
v = u + at = 0 + 4 × 10 = 40 m/s
 v2 = u2 + 2as
v2 = (0)2 + 2as
v2 = 2as
s = v2/2a = 40 × 40/ 2 × 4 = 1600/8 = 200m
Question 67.
 Name the type of motion in which speed remains constant but the velocity of the body changes.
 Name the physical quantity which changes continuously during uniform circular motion.
 Is the motion of satellites around the earth uniform or accelerated?
Answer 67.
 In accelerated motion, speed remains constant and the velocity of the body changes.
 The velocity of a particle changes continuously during uniform circular motion.
 The motion of satellites around the earth is accelerated because speed is constant and the direction of motion changes continuously.
Question 68.
 Derive the equation of motion, S = ut + ½ at2 by graphical method.
 Which of the two bodies A and B in the following graph is moving at a higher speed and why?
Answer 68:
 Suppose the object travels distance S in time t. Uniform acceleration is a.
In the graph given, the distance travelled by the object is obtained by the area enclosed within OABC under velocitytime graph AB.
Distance S = area OABC which is trapezium
= area of the rectangle OADC + area of the triangle ABD
= OA × OC + ½ ( AD × BD)
OA = u
OC = AD = t
BD = at
On substituting the values, we get
S = u × t + ½ (t × at)
S = ut + ½ at 2
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 Speed of the body is directly proportional to ϴ,
this means that a larger value has greater speed.
In the graph,
B has a higher speed compared to A.
B has a larger ϴ value.
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Question 69.
 Derive v2 – u2 = 2as using velocitytime graph.
 A ball is dropped from a height of 20 m. if its velocity increases uniformly at the rate of 10 m/s, with what velocity will it strike the ground?
Answer 69.
The velocitytime graph for uniform acceleration is given below.
To derive the third equation of motion,
s = area of the trapezium OABC
s = ½ ( sum of parallel sides) × perpendicular distance between the two parallel sides
s = ½ (OA + BC) × OC
After putting respective values in the equation we get,
S = ½ ( u + v) × t ………………(1)
We know that v = u + at
t = (v u ) / a
putting the value of t in equation 1 we get,
s = ½ (u + v) × (v – u)/a
2as = (u +v) (v – u) = v2 – u2
The final equation is 2as = v2 – u2
Hence derived.
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 u = 0
s = 20 m
a = 10 m/s 2
v = ?
v2 = u2 + 2as
v2 = 0 + 2 × 10 × 20
v2 = 400
therefore v = 20 m/s
Question 70. Study the velocitytime graph and calculate the following:
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 The acceleration from A to B
 The acceleration from B to C
 The distance covered in the region ABE
 The average velocity from C to D
 The distance covered in the region BCFE
Answer 70. (a) To calculate acceleration from A to B
a = (250)(30) = 8.3 m/s2
(b) Acceleration from B to C = a = (2025)(43) = 5m/s2
(c) Distance covered in the ABE region:
= ½ × 3 × 25
= 37.5 m
(d) Average velocity = V = (20 – 0)/2 = 10m/s
(e) Area BCFE is a trapezium.
Distance covered in BCFE = ½ (25 + 20) × (4 – 3) = 22.5 m
Question 71.
 Give one similarity and difference in the graph given below:
 What is meant by acceleration being positive and negative? Give example.
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Answer 71:
 The similarity is both the graphs are showing uniform acceleration.
The dissimilarity is
 In the first graph, u = 0. The body starts from rest.
 In the second graph, the initial velocity is nonzero.
 Positive acceleration is in the direction of velocity. For example, a bus moves at an increasing speed.
Negative acceleration, in this, acceleration is opposite to the direction of velocity. For example, the speed of a moving car suddenly decreases when a brake is applied.
Question 72.
 When is a body said to be in uniform and nonuniform acceleration?
 A train starts from rest and accelerates uniformly for 30 s to acquire a velocity of 108 km/h. It travels with this velocity for 20 min. The driver now applies breaks and the train retards uniformly to stop after 20 s. Find the total distance covered by the train.
Answer 72.
 A body is said to be in uniform acceleration when it travels in a straight path and velocity increases or decreases in equal time intervals.
When a body travels in a straight path and velocity increases or decreases in an unequal amount in an equal interval of time, it is said to be in nonuniform acceleration.
 In the given question,
The train is starting from rest.
u = 0, t = 30 s , v = 108 km/h
converting 108 km/h to m/s,
= 108 × 5/18 = 30 m/s
a = vu/t = 1m/s2
Distance s1 = ut + ½ at2 = 0 + ½ × 1 × 30 × 30 = 450m
At uniform velocity of 30 m/s for 20 min = 1200 s
Distance s2 = 30 × 1200 = 36000 m
When brake is applied,
t = 20s
u = 30 m/s
v = 0
a = v – u /t
a = 0 – 30 / 20 = 1.5 m/s 2
to calculate s3
s3 = v2 u2 / 2a = 0 – (30)2 / 2 × 1.5 = 300m
total distance = s1 + s2 + s3 = 450 + 36000 + 300 = 36750 m.
Question 73. Give examples of the following:
 A situation where the object is at rest and in motion simultaneously.
 A motion in which acceleration is nonuniform.
 A motion in which acceleration is in the direction of motion of an object.
 An example where an object moves in a certain direction with an acceleration in the perpendicular direction.
 An object with constant acceleration and zero velocity.
Answer 73.
 A passenger travelling in a bus is at rest for fellow passengers, but at the same time, it is in motion with respect to the bus.
 A car travelling on a straight road with an increase in speed in an unequal amount in equal time intervals is an example of nonuniform acceleration.
 In a freely falling object, acceleration is in the direction of motion of an object.
 When an aeroplane flies horizontally, the acceleration due to gravity acts on it in a vertically downward direction.
 When an object is thrown vertically upwards, the velocity at the highest point is zero. But the acceleration is constant, i.e. 9.8 m/s 2.
Question 74. Fill in the blanks:
 ___________ and _____________ can change the velocity of an object.
 A body in uniform circular motion has ________ speed but due to the change in the direction of motion, its ______________ changes at every point.
 The rate of change of velocity is called _______________
 When final velocity is less than initial velocity, then acceleration is __________
 ____________ is essential to describe the position of an object.
 ____________ is the simplest type of motion.
 Phenomena like the arrival of day and night indicate the _________.
 Motion and rest are never __________
Answer 74.
 Object’s speed and direction of motion can change the velocity of an object.
 A body in uniform circular motion has a constant speed, but due to the change in the direction of motion, its velocity changes at every point.
 The rate of change of velocity is called acceleration.
 When the final velocity is less than the initial velocity, then acceleration is negative.
 The reference point is essential to describe the position of an object.
 Motion in a straight line is the simplest type of motion.
 Phenomena like the arrival of day and night indicate the motion of the earth.
 Motion and rest are never absolute.
Question 75. A cyclist moving along a circular path of a radius of 63 m completes three rounds in 3 minutes. Calculate:
 Total distance covered
 Net displacement
 The speed of the cyclist
Answer 75.
 Total distance covered = s = 2πr ×t
s = 2 × 22/7 × 63 × 63
= 1188 m
 The net displacement is zero.
 Speed = distance covered / time taken
Time = 3 min = 180 s
Speed = 1188/180 = 6.6 m/s
Question 76. Give an example where motion is inferred indirectly.
Answer 76. Motion is inferred indirectly in many ways, like the movement of leaves, dust particles, and branches of trees. It can be in the form of the feeling of blowing air on the face.
Question 77. State the type of force – balanced or unbalanced, that acts on a rubber ball when we press it between our hands. Give a reason for your answer and mention the effect produced in the ball by this force.
Answer 77. A balanced force is acting on a rubber ball when it is pressed between our hands. Because an equal and opposite force is changing the shape of a ball.
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Q.1 A box is being pushed on a smooth frictionless ice rink. Choose the correct statement in the context of the motion of the box.
The box speeds up.
The box moves at a constant speed.
The box slows down.
The box does not move.
Marks:1
Ans
The box speeds up.
Q.2 Calculate the net force.
10 N
60 N
70 N
130 N
Marks:1
Ans
10 N
Q.3 An 8000 kg engine pulls a train of 5 wagons, each wagon of mass 2000 kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a friction force of 5,000 N.
Calculate
a) the net accelerating force,
b) the acceleration of the train, and
c) the force of wagon 1 on wagon 2.
Marks:5
Ans
a) Net accelerating force = 40,000 N – 5000N = 35,000 N
b) Mass of the train, m = 5—2000 = 10,000 kg
Net force acting on the train = 35000N
So, Acceleration of the train = Force / Mass
= 35000 /10000 = 3.5 m/s^{2}.
c) Force exerted by wagon 1 on wagon 2
= Net accelerating force – Force acting on wagon 1
= 35000 N – 2000 kg x 3.5 m/s^{2}
= 35000 N – 7000 N = 28000 N
Q.4 A swimmer swims forward , even though he pushes water backward while swimming. Why?
Marks:1
Ans
While swimming,a person pushes the water backward with his hand.
In reaction to this action , the water pushes the person in the forward direction with equal force.
CBSE Class 9 Science Important Questions
FAQs (Frequently Asked Questions)
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2. An object may appear to be moving to one person and resting in the other person simultaneously. Justify.
An object at rest may appear to be moving to the other person. For example, consider the object at rest like a boy standing on the ground, a tree, or a building. These objects appear to be moving to the person travelling in a moving bus in a direction opposite to that of the bus.
3. How will the equation of motion for an object moving with a constant velocity change?
In case of uniform velocity
A4: In case of uniform velocity
v = u
a = 0
v = u + at
this shows v = u
s = ut + ½ at 2
s = ut
v2 = u2 + 2as which shows v2 = u2