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Assignments Class 9 Physics Pdf Download

Students can refer to Assignments for Class 9 Physics available for download in Pdf. We have given below links to subject-wise free printable Assignments for Physics Class 9 which you can download easily. All assignments have a collection of questions and answers designed for all topics given in your latest NCERT Books for Class 9 Physics for the current academic session. All Assignments for Physics Grade 9 have been designed by expert faculty members and have been designed based on the type of questions asked in standard 9 class tests and exams. All Free printable Assignments for NCERT CBSE Class 9, practice worksheets, and question banks have been designed to help you understand all concepts properly. Practicing questions given in CBSE NCERT printable assignments for Class 9 with solutions and answers will help you to further improve your understanding. Our faculty have used the latest syllabus for Class 9. You can click on the links below to download all Pdf assignments for class 9 for free. You can get the best collection of Kendriya Vidyalaya Class 9 Physics assignments and questions workbooks below.

Class 9 Physics Assignments Pdf Download

CBSE NCERT KVS Assignments for Physics Class 9 have been provided below covering all chapters given in your CBSE NCERT books. We have provided below a good collection of assignments in Pdf for Physics standard 9th covering Class 9 questions and answers for Physics. These practice test papers and workbooks with question banks for Class 9 Physics Pdf Download and free CBSE Assignments for Class 9 are really beneficial for you and will support in preparing for class tests and exams. Standard 9th students can download in Pdf by clicking on the links below.

Subjectwise Assignments for Class 9 Physics

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Unit 1: Motion

About this unit.

Motion is all around us, from moving cars to flying aeroplanes. Motion can have different features like speed, direction, acceleration, etc. In this chapter, we will understand these features in detail and see how it can help us predict the future of these moving things.

Distance and displacement

• Distance and displacement introduction (Opens a modal)
• Distance and displacement in one dimension (Opens a modal)

Average speed and average velocity

• Average speed & velocity (with examples) (Opens a modal)
• Calculating average velocity or speed (Opens a modal)
• Average speed for entire journey - solved numerical (Opens a modal)
• Average velocity and speed in one direction: word problems 4 questions Practice
• Average velocity and speed with direction changes: word problems 4 questions Practice

Instantaneous speed and velocity

• Instantaneous speed & velocity (Opens a modal)

Acceleration

• Acceleration (Opens a modal)
• Airbus A380 take-off time (Opens a modal)
• Acceleration and velocity 7 questions Practice

Position time graphs

• Position-time graphs (Opens a modal)
• Calc. velocity from position time graphs (Opens a modal)

Velocity time graphs

• Velocity vs. time graphs (Opens a modal)
• Calculating displacement from v-t graphs (Opens a modal)

Deriving equations of motion

• Deriving 3 equations of motion (from v-t graph) (Opens a modal)

Problem solving using kinematic equations

• Using equations of motion (1 step numerical) (Opens a modal)
• Using equations of motion (2 steps numerical) (Opens a modal)
• Kinematic equations: calculations 4 questions Practice

Uniform circular motion

• Calc. speed & time in a uniform circular motion - Solved numerical (Opens a modal)

• NCERT Solutions
• NCERT Class 9
• NCERT 9 Science
• Chapter 8: Motion

NCERT Solutions for Class 9 Science Chapter 8: Motion

Ncert solutions class 9 science chapter 8 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.

NCERT Solutions for Class 9 Science Chapter 8 Motion is designed with the intention of clarifying doubts and concepts easily. Class 9 NCERT Solutions in Science is a beneficial reference and guide that helps students clear doubts instantly in an effective way. 

Download Exclusively Curated Chapter Notes for Class 9 Science Chapter – 8 Motion

Download most important questions for class 9 science chapter – 8 motion.

NCERT Solutions for Class 9 Science approaches students in a student-friendly way and is loaded with questions, activities, and exercises that are CBSE exam and competitive exam-oriented. NCERT Solutions for Class 9 Science is the contribution of our faculty, having vast teaching experience. It is developed keeping in mind the concept-based approach along with the precise answering method for CBSE examinations. Refer to NCERT Solutions for Class 9 for best scores in CBSE and competitive exams. It is a detailed and well-structured solution for a solid grip on the concept-based learning experience. NCERT for Class 9 Science Solutions is made available in both web and PDF format for ease of access.

• Chapter 1 Matter in Our Surroundings
• Chapter 2 Is Matter Around Us Pure?
• Chapter 3 Atoms and Molecules
• Chapter 4 Structure of the Atom
• Chapter 5 The Fundamental Unit of Life
• Chapter 6 Tissues
• Chapter 7 Diversity in Living Organisms
• Chapter 9 Force And Laws Of Motion
• Chapter 10 Gravitation
• Chapter 11 Work and Energy
• Chapter 12 Sound
• Chapter 13 Why Do We Fall Ill?
• Chapter 14 Natural Resources
• Chapter 15 Improvement in Food Resources

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Access Answers of Science NCERT class 9 Chapter 8: Motion  (All intext and exercise questions solved)

Intext Questions – 1   Page: 100

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

Yes, an object which has moved through a distance can have zero displacement if it comes back to its initial position.

Example: If a person jogs in a circular park which is circular and completes one round. His initial and final position is the same.

Hence, his displacement is zero.

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

Side of the given square field = 10m

Hence, the perimeter of a square = 40 m

Time taken by the farmer  to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of 1 m

Distance covered by the farmer in 2 min 20 sec = 1 x 140 = 140 m

The total number of rotations taken by the farmer to cover a distance of 140 m = total distance/perimeter

At this point, let us say the farmer is at point B from the origin O

Therefore,  from Pythagoras theorem, the displacement s = √(10 2 +10 2 )

s =  10 √ 2

s = 14.14 m

3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.

Neither of the statements is true.

(a) Given statement is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero.

(b) Given statement is false because the displacement of an object can be equal to, but never greater than the distance travelled.

Intext Questions – 2   Page: 102

1. Distinguish between speed and velocity.

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

Since average speed is the total distance travelled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance travelled is equal to the displacement.

3. What does the odometer of an automobile measure?

An odometer, or odograph, is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.

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

The path of an object in uniform motion is a straight line.

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, 3 × 10 8 m/s.

Given that the signal travels in a straight line, the distance between the spaceship and the ground station is equal to the total distance travelled by the signal.

5 minutes = 5*60 seconds = 300 seconds.

Speed of the signal = 3 × 10 8 m/s.

Therefore, total distance = (3 × 10 8 m/s) * 300s

= 9*10 10 meters.

Intext Questions – 3   Page: 103

1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Uniform Acceleration:  When an object is travelling in a straight line with an increase in velocity at equal intervals of time, then the object is said to be in uniform acceleration.

The free-falling of an object is an example of uniform acceleration.

Non-Uniform Acceleration:  When an object is travelling with an increase in velocity but not at equal intervals of time is known as non-uniform acceleration.

Bus moving or leaving from the bus stop is an example of non-uniform acceleration.

2. A bus decreases its speed from 80 km h –1 to 60 km h –1 in 5 s. Find the acceleration of the bus.

Given, the initial velocity (u) = 80km/hour = 80000m/3600s= 22.22 m.s -1

The final velocity (v) = 60km/hour = 60000m/3600s= 16.66 m.s -1

Time frame, t = 5 seconds.

Therefore, acceleration (a) =(v-u)/t = (16.66 m.s -1 – 22.22 m.s -1 )/5s

= -1.112 m.s -2

Therefore, the total acceleration of the bus is -1.112m.s -2 . It can be noted that the negative sign indicates that the velocity of the bus is decreasing.

3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h –1 in 10 minutes. Find its acceleration.

Given parameters

Initial velocity (u) = 0

Final velocity (v) = 40 km/h

v = 40 × (5/18)

v = 11.1111 m/s

Time (t) = 10 minute

t = 60 x 10

Acceleration (a) =?

Consider the formula

11.11 = 0 + a × 600

11,11 = 600 a

a = 11.11/600

a = 0.0185 ms -2

Intext Questions – 4   Page: 107

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

For uniform motion, the distance-time graph is a straight line. On the other hand, the distance-time graph of an object in non-uniform motion is a curve.

The first graph describes the uniform motion and the second one describes the non-uniform motion.

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

The distance-time graph can be plotted as follows.

When the slope of the distance-time graph is a straight line parallel to the time axis, the object is at the same position as time passes. That means the object is at rest.

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

The speed-time graph can be plotted as follows.

Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion.

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

Considering an object in uniform motion, its velocity-time graph can be represented as follows.

Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by OA*OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:

Area under the velocity-time graph = velocity*time.

Substituting the value of velocity as displacement/time in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.

Intext Questions – 5 Page: 109,110

1. A bus starting from rest moves with a uniform acceleration of 0.1 m s -2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

(a) Given, the bus starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) = 0.1 m.s -2

Time = 2 minutes = 120 s

Acceleration is given by the equation a=(v-u)/t

Therefore, terminal velocity (v) = (at)+u

= (0.1 m.s -2 * 120 s) + 0 m.s -1

= 12 m.s -1 + 0 m.s -1

Therefore, terminal velocity (v) = 12 m/s

(b) As per the third motion equation, 2as = v 2 – u 2

Since a = 0.1 m.s -2 , v = 12 m.s -1 , u = 0 m.s -1 , and t = 120 s, the following value for s (distance) can be obtained.

Distance, s =(v 2 – u 2 )/2a

=(12 2 – 0 2 )/2(0.1)

Therefore, s = 720 m.

The speed acquired is 12 m.s -1 and the total distance travelled is 720 m.

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

Given, initial velocity (u) = 90 km/hour = 25 m.s -1

Terminal velocity (v) = 0 m.s -1

Acceleration (a) = -0.5 m.s -2

As per the third motion equation, v 2 -u 2 =2as

Therefore, distance traveled by the train (s) =(v 2 -u 2 )/2a

s = (0 2 -25 2 )/2(-0.5) meters = 625 meters

The train must travel 625 meters at an acceleration of -0.5 ms -2 before it reaches the rest position.

3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s -2 . What will be its velocity 3 s after the start?

Given, initial velocity (u) = 0 (the trolley begins from the rest position)

Acceleration (a) = 0.02 ms -2

Time (t) = 3s

As per the first motion equation, v=u+at

Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms -2 )(3s)= 0.06 ms -1

Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s -1

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

Given, the car is initially at rest; initial velocity (u) = 0 ms -1

Acceleration (a) = 4 ms -2

Time period (t) = 10 s

As per the second motion equation, s = ut+1/2 at 2

Therefore, the total distance covered by the car (s) = 0 * 10m + 1/2 (4ms -2 )(10s) 2

= 200 meters

Therefore, the car will cover a distance of 200 meters after 10 seconds.

5. A stone is thrown in a vertically upward direction with a velocity of 5 m s -1 . If the acceleration of the stone during its motion is 10 m 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?

Given, initial velocity (u) = 5 m/s

Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)

Acceleration = 10 ms -2 in the direction opposite to the trajectory of the stone = -10 ms -2

As per the third motion equation, v 2 – u 2 = 2as

Therefore, the distance travelled by the stone (s) = (0 2 – 5 2 )/ 2(10)

Distance (s) = 1.25 meters

As per the first motion equation, v = u + at

Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a

=(0-5)/-10 s

Time taken = 0.5 seconds

Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.

Exercises Page: 112,113

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

Given, diameter of the track (d) = 200m

Therefore, the circumference of the track (π*d) = 200π meters.

Distance covered in 40 seconds = 200π meters

Distance covered in 1 second = 200π/40

Distance covered in 2minutes and 20 seconds (140 seconds) = 140 * 200π/40 meters

= (140*200*22)/(40* 7) meters = 2200 meters

Number of rounds completed by the athlete in 140 seconds = 140/40 = 3.5

Therefore, the final position of the athlete (with respect to the initial position) is at the opposite end of the circular track. Therefore, the net displacement will be equal to the diameter of the track, which is 200m.

Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Given, distance covered from point A to point B = 300 meters

Distance covered from point A to point C = 300m + 100m = 400 meters

Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds

Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds

Displacement from A to B = 300 meters

Displacement from A to C = 300m – 100m = 200 meters

Average speed = total distance travelled/ total time taken

Average velocity = total displacement/ total time taken

Therefore, the average speed while traveling from A to B = 300/150 ms -1 = 2 m/s

Average speed while traveling from A to C = 400/210 ms -1 = 1.9 m/s

Average velocity while traveling from A to B =300/150 ms -1 = 2 m/s

Average velocity while traveling from A to C =200/210 ms -1 = 0.95 m/s

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km.h –1 . On his return trip along the same route, there is less traffic and the average speed is 30 km.h –1 . What is the average speed for Abdul’s trip?

Distance travelled to reach the school = distance travelled to reach home = d (say)

Time taken to reach school = t 1

Time taken to reach home = t 2

therefore, average speed while going to school = total distance travelled/ total time taken = d/t 1 = 20 kmph

Average speed while going home = total distance travelled/ total time taken = d/t 2 = 30 kmph

Therefore, t 1 = d/20 and t 2 = d/30

Now, the average speed for the entire trip is given by total distance travelled/ total time taken

= (d+d)/(t 1 +t 2 )kmph = 2d/(d/20+d/30)kmph

= 2/[(3 + 2)/60]

= 120/5 kmh -1 = 24 kmh -1

Therefore, Abdul’s average speed for the entire trip is 24 kilometers per hour.

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

Given, initial velocity of the boat = 0 m/s

Acceleration of the boat = 3 ms -2

Time period = 8s

As per the second motion equation, s = ut + 1/2 at 2

Therefore, the total distance travelled by boat in 8 seconds = 0 + 1/2 (3)(8) 2

= 96 meters

Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.

5. A driver of a car travelling at 52 km h –1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h –1 in another car applies his brakes slowly and stops in 10 s. 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?

The speed v/s time graphs for the two cars can be plotted as follows.

The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.

Therefore, displacement of the first car = area of triangle AOB

= (1/2)*(OB)*(OA)

But OB = 5 seconds and OA = 52 km.h -1 = 14.44 m/s

Therefore, the area of the triangle AOB is given by: (1/2)*(5s)*(14.44ms -1 ) = 36 meters

Now, the displacement of the second car is given by the area of the triangle COD

= (1/2)*(OD)*(OC)

But OC = 10 seconds and OC = 3km.h -1 = 0.83 m/s

Therefore, area of triangle COD = (1/2)*(10s)*(0.83ms -1 ) = 4.15 meters

Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters. Therefore, the first car (which was traveling at 52 kmph) travelled farther post the application of brakes.

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

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

(a) Since the slope of line B is the greatest, B is travelling at the fastest speed.

(b) Since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.

(c) Since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.

Since the initial point of an object C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 km

When B passes A, the distance between the origin and C is 8km

Therefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km

(d) The distance that object B has covered at the point where it passes C is equal to 9 graph units.

Therefore, total distance travelled by B when it crosses C = 9*(4/7) = 5.14 km

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

Given, initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance travelled by the ball (s) = 20m

Acceleration (a) = 10 ms -2

As per the third motion equation,

v 2 – u 2 = 2as

= 2*(10ms -2 )*(20m) + 0

v 2 = 400m 2 s -2

Therefore, v= 20ms -1

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation,

Therefore, t = (v-u)/a

= (20-0)ms -1 / 10ms -2

= 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

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

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

The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

(1/2)*4*6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6 th to the 10 th second.

9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is possible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

Circular motion is an example of an object moving with acceleration but with uniform speed.

An object moving in a circular path with uniform speed is still under acceleration because the velocity changes due to continuous changes in the direction of motion.

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

10. 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.

Given, the radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2*π*42250km = 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.

NCERT Class 9 Science Chapter 8 explains the concept of motion, and types of motion with relevant examples for a clear understanding of the concept. It explains the causes of phenomena like sunrise, sunset, and changing of the seasons. It helps students understand uniform and non-uniform motion. Distance-time graphs and velocity-time graphs, which are considered important concepts for examination, are explained in an easy way in NCERT Solutions . It describes how the acceleration of an object is the change in velocity per unit time. NCERT Class 9 Science Chapter 8 is covered under Unit III: Motion, Force and Work and can get you maximum marks.

• NCERT Solutions for Class 9 explains motion in terms of distance moved or displacement.
• Uniform and non-uniform motions of objects are explained through the graph and examples.
• Uniform circular motion concept is made understandable in a simple way.
• Problems on acceleration, velocity, and average velocity are also solved.

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

• A simple and easily understandable approach is followed in NCERT Solutions to make students aware of topics.
• Provides complete solutions to all the questions present in the respective NCERT textbooks.
• NCERT Solutions offers detailed answers to all the questions to help students in their preparations.
• These solutions will be useful for CBSE exams, Science Olympiads, and other competitive exams.

Disclaimer:

Dropped Topics –  8.5 Equations of motion by graphical method, 8.5.1 Equation for Velocity–Time Relation, 8.5.2 Equation for Position–Time relation and 8.5.3 Equation for Position– Velocity.

Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 8

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CBSE Worksheets for Class 9 Physics

CBSE Worksheets for Class 9 Physics: One of the best teaching strategies employed in most classrooms today is Worksheets. CBSE Class 9 Physics Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and problem-solving capabilities. So in order to help you with that, we at WorksheetsBuddy have come up with Kendriya Vidyalaya Class 9 Physics Worksheets for the students of Class 9. All our CBSE NCERT Class 9 Physics practice worksheets are designed for helping students to understand various topics, practice skills and improve their subject knowledge which in turn helps students to improve their academic performance. These chapter wise test papers for Class 9 Physics will be useful to test your conceptual understanding.

Board: Central Board of Secondary Education(www.cbse.nic.in) Subject: Class 9 Physics Number of Worksheets: 25

CBSE Class 9 Physics Worksheets PDF

All the CBSE Worksheets for Class 9 Physics provided in this page are provided for free which can be downloaded by students, teachers as well as by parents. We have covered all the Class 9 Physics important questions and answers in the worksheets which are included in CBSE NCERT Syllabus. Just click on the following link and download the CBSE Class 9 Physics Worksheet. CBSE Worksheets for Class 9 Physics can also use like assignments for Class 9 Physics students.

• CBSE Worksheets for Class 9 Physics All Chapters Assignments 
• CBSE Worksheets for Class 9 Physics Force and Laws of Motion Assignment 1
• CBSE Worksheets for Class 9 Physics Force and Laws of Motion Assignment 2
• CBSE Worksheets for Class 9 Physics Force and Laws of Motion Assignment 3
• CBSE Worksheets for Class 9 Physics Gravitation Assignment 1
• CBSE Worksheets for Class 9 Physics Gravitation Assignment 2
• CBSE Worksheets for Class 9 Physics Gravitation Assignment 3
• CBSE Worksheets for Class 9 Physics Motion Assignment 1
• CBSE Worksheets for Class 9 Physics Motion Assignment 2
• CBSE Worksheets for Class 9 Physics Sound Assignment
• CBSE Worksheets for Class 9 Physics Work & Energy Assignment 1
• CBSE Worksheets for Class 9 Physics Work & Energy Assignment 2
• CBSE Worksheets for Class 9 Physics Assignment 1
• CBSE Worksheets for Class 9 Physics Assignment 2
• CBSE Worksheets for Class 9 Physics Assignment 3
• CBSE Worksheets for Class 9 Physics Assignment 4
• CBSE Worksheets for Class 9 Physics Assignment 5
• CBSE Worksheets for Class 9 Physics Assignment 6
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• CBSE Worksheets for Class 9 Physics Assignment 8
• CBSE Worksheets for Class 9 Physics Assignment 9
• CBSE Worksheets for Class 9 Physics Assignment 10
• CBSE Worksheets for Class 9 Physics Assignment 11
• CBSE Worksheets for Class 9 Physics Assignment 12
• CBSE Worksheets for Class 9 Physics Assignment 13

Advantages of CBSE Class 9 Physics Worksheets

• By practising NCERT CBSE Class 9 Physics Worksheet , students can improve their problem solving skills.
• Helps to develop the subject knowledge in a simple, fun and interactive way.
• No need for tuition or attend extra classes if students practise on worksheets daily.
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Physics Numerical for class 9 motion & other chapters with answers & [PDF]

Last updated on December 12th, 2023 at 01:53 pm

This is an online guide on Physics Numerical for Class 9 . It presents a super-set of Numerical Problems on Motion and other Chapters in Class 9 Science for CBSE, ICSE, & UPSC. The full set contains 101+ problems with solutions for most of the problems.

For faster loading of the page, we have broken down the series into multiple pages. The links are given at the end of this post for the next pages. You will get a PDF download link as well on this page.

Numerical Problems from class 9 Motion (with solution)

Physics numericals class 9 from different chapters.

The next set of numerical questions (set 1) covers questions from the Motion chapter (kinematics) of class 9 physics and these can be easily solved using motion equations ( Suvat equations ).

Formulas Used

v=u + at s= ut + (1/2)at 2 s = [(u+v)/2]. t v 2 =u 2 + 2as

1) A speedboat has an acceleration of 2 m/s 2 . What would the final velocity of the speedboat be after 5 seconds if the initial velocity of the speedboat is 4 m/s?

a = 2 m/s 2 u = 4 m/s t = 5 s final velocity = v v = u + at = 4 + 2×5 = 14 m/s

2) A vehicle has an acceleration of 5 m/s 2 . What would the final velocity of the vehicle be after 10 seconds if the initial velocity of the vehicle is 20 m/s? What is the displacement of the vehicle during the 10 s time interval?

a = 5 m/s 2 u = 20 m/s t = 10 s final velocity = v v = u + at = 20 + 5×10 = 70 m/s final velocity = 70 m/s Displacement during the 10 s time interval = s s = ut + (1/2)at 2 => s = 20×10 + (½)x5x10 2 => s= 200 + 250 = 450 m Displacement during the 10 s time interval = 450 m

3) A child on a toboggan starts from rest and accelerates down a snow-covered hill at 0.8 m/s 2 . How long does it take the child to reach the bottom of the hill if it is 25.0 m away?

Solution: initial velocity u = 0 acceleration a = 0.8 m/s 2 distance = s = 25 m time taken t = ? s= ut + (1/2)at 2 As here, u = 0, so we can write: s= (1/2)at 2 => t 2 =(2 s /a ) = (2×25)/0.8 = 62.5 time t = 7.9 sec

4. A car accelerates uniformly from a velocity of 21.8 m/s to a velocity of 27.6 m/s. The car travels 36.5 m during this acceleration.

a) What was the acceleration of the car?

b) Determine the time interval over which this acceleration occurred.

a) final velocity v = 27.6 m/s initial velocity u = 21.8 m/s distance traveled s=36.5 m We will use this equation: v 2 = u 2 + 2as a = ( v 2 – u 2 ) /(2s) = (27.6 2 – 21.8 2 )/(2×36.5) = 3.92 m/s 2 Acceleration = 3.92 m/s 2 b) time interval over which this acceleration occurred = t v = u + at => t = (v-u)/a = (27.6-21.8) / 3.92 = 1.48 seconds.

5. A motorcycle starts from rest and accelerates at +2.50 m/s 2 for a distance of 150.0 m. It then slows down with an acceleration of –1.50 m/s 2 until the velocity is +10.0 m/s. Determine the total displacement of the motorcycle.

First part: displacement = s = 150 m We need to find out the final velocity(v) of this accelerating part. v 2 = u 2 + 2as => v 2 = 0 + 2×2.5×150 = 750 => v = 27.4 m/s This velocity becomes the initial velocity of the second part of the journey. Second part: u = 27.4 m/s a = -1.5 m/s 2 v= 10 m/s v 2 = u 2 + 2as 10 2 = 27.4 2 + 2.(-1.5).s => 100 = 750 – 3s s= 650/3 = 216.7 m Second part displacement = 216.7 m Considering a straight-line journey without a change of direction, total displacement = (150 + 216.7) m = 366.7 m

[PDF] on Motion Numericals download

So let’s start with sets of Physics Numericals for Class 9 which contains Numerical problems & questions from various chapters of Class 9 physics with solutions & answers.

Problems based on Newton’s Second Law

F = ma m = F/a a = F/m F = (mv – mu)/t

1) A force of 1000 Newton is applied on a 25 kg mass for 5 seconds. What would be its velocity?

  Solution: F =1000 N m=25kg t= 5 sec u=0 v=? Acceleration = a = F/m = 1000/25 = 40 m/s 2  v = u + at =0 + 40X5 = 200 m/s

2) A force is applied to a mass of 16 kg for 3 seconds. As the force is removed the mass moves 81 meters in 3 seconds. What was the value of the force applied?

Solution: First 3 secs: Force is present, and in the next 3 secs Force is not there, so uniform velocity for the last 3 seconds is mentioned. In the last 3 seconds, the mass travels with uniform velocity (Where no force means there is no acceleration). So the velocity in the next 3 secs: V = 81/3 = 27 m/s This velocity was attained in the first 3 seconds when the force was present. So in the first 3 seconds, velocity changes from 0 to 27m/s. Therefore the acceleration in the first 3 seconds caused by the force present = (27-0)/3 = 9 m/s². So the force applied on the mass for first 3 secs= mass x acc = 16 x 9 = 144 N

3) A mass of 50 kg was moving with a velocity of 400 m/s. A force of 40000 N is applied to the mass and its velocity is reduced to 50 m/s after some time. What is the distance traveled by the mass during this period?

Solution: Mass = 50 kg u = 400 m/s v= 50 m/s F = 40000 N t = unknown Distance traveled in time t =? Acceleration = a = F/m = 40000/50 = 800 m/s 2 Again, acceleration a = change of velocity /t or, t = change of velocity/a = 350/800 sec Distance traveled: s = ut – (1/2)a t 2 = 400 * (350/800) – (1/2) 800. (350/800) 2 = 175 – 76.5 = 98.5 m

4) A ball of 150 g mass moves with 12 m/s and bounces back after hitting a wall with 20 m/s after a small duration of 0.01 seconds. What was the force applied to the ball when it hits the wall?

Solution: Mass = 150 gram = 150/1000 kg = 0.15 kg Change of velocity = 20 – (-12) = 32 m/s T = 0.01 sec Acceleration = change in velocity / time = 32/0.01 = 3200 m/s 2 Force = mass X Acceleration = 0.15 X 3200 N= 480 N.

5) Momentum of an object changes from 100 kg m/s to 200 kg m/s in 2 seconds. What is the force applied to it?

Solution: Note: Force = ma = m(v-u)/t = (mv – mu)/t = change of momentum /time = rate of change of momentum Force= change of momentum /time = (200-100)/2 = 100/2 = 50 N

6) A force of 200 N is applied to a body and its velocity changes from 5m/s to 10m/s in a second. What is the mass of the body?

Solution: mass = force/acceleration here acceleration =a = (10-5)/1 = 5 m/s 2 so, mass  = 200/5 kg = 40 kg

7) A mass of 1 kg is moving from east to west with a velocity of 10 m/s. A force is applied to it for 2 seconds and its velocity becomes 5 m/s. What is the value and direction of the force?

Solution: Mass = 1kg velocity = 10 m/s (east to west) Because of the force, the velocity reduces to 5 m/s in 2 secs. acceleration = change of vel/time=5/2 = 2.5 m/s 2 Force = mass X acceleration= 1 X 2.5 = 2.5 N. As the velocity reduces because of force, (negative acc i.e. retardation) hence the direction of the force would be opposite to the initial direction. So the direction of the force will be west to east.

Problems based on the Universal Law of Gravitation

F = GMm/r 2

8) If the distance between 2 objects is doubled then how will the gravitational force between them change?

Solution: F1 = GMm/r 2 F2 = GMm/(2r) 2 = GMm/4r 2 so F2/F1 = ¼ , so the final gravitational force will be ¼th of the initial gravitational force

9) If the distance between 2 objects is halved then how will the gravitational force between them change?

Solution: F1 = GMm/r 2 F2 = GMm/(r/2) 2 =4 GMm/r 2 so F2/F1 = 4/1 so the final gravitational force will be 4 times the initial gravitational force

10) If the masses of the 2 objects are doubled then how will the gravitational force between them change?

Solution: F1 = GMm/r 2 F2 = (G2M2m)/(r) 2 =4 GMm/r 2 so F2/F1 = 4/1 so the final gravitational force will be 4 times the initial gravitational force

Problems based on Force, Acceleration, and Motion

11) A force acts for 10 s on a stationary body of mass 100 kg after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate (i) the velocity acquired by the body, (ii) the acceleration produced by the force, and (iii) the magnitude of the force. [ Solution: Q11 ]

12) A force acts for 0.1 sec on a body of mass 2 kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2 m/s. Find the magnitude of the force. Solution: solution Q12

13) A cricket ball of mass 100 g moving at a speed of 30 m/s is brought to rest by a player in 0.03 s. find the average force applied by the player. Solution: solution Q13

14) A body of mass 500 g, initially at rest, is acted upon by a force that causes it to move a distance of 4 m in 2 sec. Calculate the force applied. Solution: solution Q14

15) A force causes an acceleration of 10 m/s^2 in a body of mass 500 g. What acceleration would be caused by the same force in a body of mass 5 kg. Solution: solution Q15

16) A bullet of mass 50 g moving with an initial velocity of 100 m/s, strikes a wooden block and comes to rest after penetrating a distance 2 cm in it. Calculate: (i) Initial momentum of the bullet (ii)final momentum of the bullet (iii) retardation caused by the wooden block, and (iv) resistive force exerted by the wooden block Solution: solution Q16

Find another set of Physics Numerical for Class 9 Motion.

17) From the velocity-time graph , find out the distance traveled in each section A, B, C, and D. Also find the total distance traveled.

Solution: solution Q17

18) A car starts from rest and accelerates uniformly over a time of 7 seconds for a distance of 98 m. Determine the acceleration of the car. Solution: solution Q18

19) A car traveling at 20 m/s skids to a stop in 3 s. Determine the skidding distance of the car (assume uniform acceleration). Solution: solution Q19 20) An airplane accelerates down a runway at 4 m/s^2 for 30 s until it finally lifts off the ground. Determine the distance traveled before takeoff. Solution: solution Q20

21) A feather is dropped on the moon from a height of 10 meters. The acceleration of gravity on the moon is one-sixth of that on the earth. Determine the time for the feather to fall to the surface of the moon. Solution: solution Q 21

22) If a rocket-powered sled is accelerated to a speed of 400 m/s in 2 seconds, then what is the acceleration, and what is the distance that the sled travels? Solution: solution Q 22

23) An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s 2 . The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway? Solution: solution Q 23

24) Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall? Solution: solution Q 24

25) A bike accelerates uniformly from rest to a speed of 10 m/s over a distance of 30 m. Determine the acceleration of the bike. Solution: solution Q25

26) A race car accelerates uniformly from 25 m/s to 50 m/s in 5 seconds. Determine the acceleration of the car and the distance traveled. Solution: solution Q 26

27) A bullet leaves a rifle with a muzzle velocity of 600 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.8 m. Determine the acceleration of the bullet (assume uniform acceleration). Solution: solution Q 27

28) A bullet is moving at a speed of 350 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.05 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.) Solution: solution Q 28

29) A plane has a takeoff speed of 100 m/s and requires 1500 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed. Solution: solution Q 29

30)A baseball is popped straight up into the air and has a hang-time of 6 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the max height is one-half the total hang time.) Solution: solution Q30

31) If a box is pushed horizontally on the floor with a force of 10 N and it moves 5 meters along the line of action of the force, then what is the work done by the Gravity or earth’s gravitational pull on that box? Solution: solution Q31

Related & Suggested Reading for the readers: 1) physics Questions & Answers for class 9 CBSE, ICSE, state 2) Class 9 Numericals set 2 on Laws of Motion and Force – hand-picked problems, Author Name Tweet

32) A train is moving at a velocity of 25 ms-1. It is brought to rest by applying the brakes which produce uniform retardation of 0.5 ms-2. Calculate (i) the velocity of the train after 10 s (ii) If the mass of the train is 20000 kg then calculate the force required to stop the train Solution: solution Q3 2

33) A spring balance is used to find the weight of a body X on the surface of the moon. The mass of the body X is 2 kg and its weight is recorded as 3.4 N. The weight of another body Y recorded by the same balance is found to be 7.65 N. Calculate the mass of the body Y. Solution: solution Q3 3

Gravitational force numerical

34) Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively kept at a distance of 10 m from one another. (Gravitational constant, G = 6.7 × 10 -11 Nm^2/kg^2 ) Solution: solution Q34

35) What is the force of gravity on a body of mass 150 kg lying on the surface of the earth? (Mass of earth = 6 × 10^24 kg; Radius of earth = 6.4 × 10^6 m; G = 6.7 × 10 -11 Nm^2/kg^2) Solution: solution Q35

36) The mass of sun is 2 × 10^30 kg and the mass of earth is 6 × 10^24 kg. If the average distance between the sun and the earth is 1.5 × 10^8 km, calculate the force of gravitation between them. Solution: solution Q36

Vertical Motion questions

The following set on physics numerical for class motion specifically covers vertical motion questions.

37) A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone is 9.8 m/s^2 directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards. Solution: solution Q37

38) A stone falls from a building and reaches the ground 2.5 seconds later. How high is the building? ( g = 9.8 m/s^2) Solution: solution Q3 8

39) A stone is dropped from a height of 20 m.( i ) How long will it take to reach the ground? ( ii ) What will be its speed when it hits the ground? ( g = 10 m/s^2) Solution: solution Q3 9

40) A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall? ( g = 9.8 m/s^2) Solution: solution Q 40

41) When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 meters. ( a ) What was the initial speed of the ball? ( b ) How much time is taken by the ball to reach the highest point? ( g =10 m/ s^2)

Solution: solution Q 41

Numerical questions on Mass, Weight & acceleration due to gravity (g)

42) A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration? ( g = 9.8 m/ s^2). Solution: solution Q42

43) A stone resting on the ground has a gravitational force of 20 N acting on it. What is the weight of the stone? What is its mass? ( g = 10 m/s^2). Solution: solution Q43

44) (a) An object has a mass of 20 kg on Earth . What will be its ( i ) mass, and ( ii ) weight, on the moon? ( g on moon = 1.6 m/s^2). ( b ) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 10^22 kg and a radius of 1.74 × 10^6 m ( G = 6.7 × 10 -11 Nm^2/kg^2). Solution: solution Q44

45) The mass of a planet is 6 × 10^24 kg and its diameter is 12.8 × 10^3 km. If the value of the gravitational constant is 6.7 × 10 -11 Nm2/kg2, calculate the value of acceleration due to gravity on the surface of the planet. Solution: solution Q45

46) The values of g at six distances A, B, C, D, E, and F from the surface of the earth are found to be 3.08 m/s^2, 9.23 m/s^2, 0.57 m/s^2, 7.34 m/s^2, 0.30 m/s^2 and 1.49 m/s^2, respectively. ( a ) Arrange these values of g according to the increasing distances from the surface of the earth (keeping the value of g nearest to the surface of the earth first) ( b ) If the value of distance F be 10000 km from the surface of the earth( hint: check if possible or not ), state whether this distance is deep inside the earth or high up in the sky. Give the reason for your answer of the place of observation Solution: solution Q46

47) A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m/s^2 and speed of sound = 330 m/s, find (i) the depth of the water surface, and (ii) the time when the echo is heard after the pebble is dropped. **[answer Hint: 2000 m ; 26.06 secs ] Solution: solution Q4 7

48) A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19·6 m/s. The ball reaches the ground after 5 s. Calculate: (i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g = 9·8 m/s^2 **[answer Hint: 24.5 m ; 29.4 m/s ] Solution: solution Q4 8

We have 53 more numerical questions in this series for class 9 physics – these are listed on the next pages of this series. You will find the links to these pages below.

6 thoughts on “ Physics Numerical for class 9 motion & other chapters with answers & [PDF] ”

I find this collection of physics problems very useful. Our class teacher has also prescribed us to solve these problems. Thank you.

Many thanks for putting in the effort for students like me. I am solving these problems even for the competitive exams I will have soon.

You have a good amount of materials for physics problems. I found this site very helpful. One request: can you post chemistry problems as well?

Thanks for your comments and suggestion, Shilpa. We will certainly plan for it.

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• Motion Class 9 Notes CBSE Science Chapter 8 (Free PDF Download)
• Revision Notes

CBSE Class 9 Science Chapter 8 - Motion Revision Notes - Free PDF Download

The theories and principles taught in NCERT chapter 8 of motion for class 9 are explained in motion chapter class 9 notes. When the position of a body changes in reference to a stationary object, it is said to be in motion. Understanding higher-level physics subjects requires a comprehension of motion. There are several types of motions, and the phenomenon of motion is regulated by definite principles. As part of the Class 9 science chapter 8 notes, we will go over all of this in great detail.

Vedantu also provides free NCERT Solutions to all the students. You can download Class 9 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations. 

Download free NCERT Solutions Class 9 Maths to amp up your preparations and to score well in your examinations.

Download CBSE Class 9 Science Revision Notes 2024-25 PDF

Also, check CBSE Class 9 Science revision notes for All chapters:

Access CBSE Class 9 Science Chapter 8 - Motion Notes

Introduction: 

One of the most common phenomena in the physical world is motion. Mechanics is the branch of Physics that deals with the behavior of moving objects.

Mechanics is divided further into two sections: Kinematics and Dynamics.

Kinematics is the study of motion without regard for the cause of motion.

Dynamics is concerned with the source of motion, which is force. 

Motion and Rest: 

An object is said to be in motion if its position in relation to its surroundings changes in a given time.

An object is said to be at rest if its position in relation to its surroundings does not change.

A frame of reference is another object or scene against which we compare the position of an object.

Take a look at the numbers. Figure 1 shows the car to the right of the tree. Figure 2 shows the car to the left of the tree after 2 seconds. The car must have moved from one location to another because the tree does not move. As a result, the tree serves as the frame of reference in this case. 

Types of Motion: 

There are three types of motion: 

Translatory motion 

Rotatory motion 

Vibratory motion 

Translatory Motion: 

A particle in translatory motion moves from one point in space to another. This movement may be in a straight line or in a curved path.

Rectilinear motion is defined as motion along a straight line.

Curvilinear motion is defined as movement along a curved path.

As an example, consider a car driving down a straight road.

Rectilinear Motion: 

Example: A car negotiating a curve 

Curvilinear Motion: 

Rotatory Motion: 

The particles of the body describe concentric circles around the axis of motion in rotatory motion.

Vibrational Motion:

Particles in vibratory motion move back and forth around a fixed point.

Distance and Displacement: 

The distance between termini A and B is 150 kilometers. A bus connects Terminus A and Terminus B. The bus travels a distance of 150 kilometers. The bus returns from terminus B to terminus A along the same route. As a result, the total distance traveled by the bus from A to B and then from B to A is 150 km + 150 km = 300 km. 

A bus traveling from point A to point B and back again.

The distance traveled by a moving object is the length of the path the object takes.

The measure of distance is a scalar quantity. The meter is the SI unit of distance.

The bus's position changed when it moved from Terminus A to Terminus B. The distance between A and B is 150 kilometers. The distance traveled on the return trip is also 150 kilometers.

Displacement is the shortest path covered by a moving object in a specified direction from the point of reference (the initial position of the body).

Note:  

However, the displacement when the bus moves from A B to B B is zero. The meter is the SI unit of displacement.

Displacement is a vector, which means that it is represented by a number with appropriate units and direction.

To emphasize the distinction between displacement and distance, consider a few more examples.

Assume a person moves 3 meters from point A to point B and 4 meters from point B to point C, as shown in the figure. He has travelled a total distance of 7 meters. But is he really 7 meters away from his starting point? No, he is only 5 meters away from his initial position, implying that he is displaced by the shortest distance between his initial and final positions.

Distance and Displacement  

To determine the displacement in this example, we can use Pythagoras' theorem. Consider an object that is changing its position with respect to a fixed point known as the origin 0. 

$\mathrm{x}_{\mathrm{i}}$ and $\mathrm{x}_{\mathrm{f}}$ are the initial and final positions of the object. Then the displacement of the

object $=x_{f}-x_{i}$

Suppose the object is travelling from $+1$ to $+4$, then displacement

$=\mathrm{x}_{\mathrm{f}}-\mathrm{x}_{\mathrm{i}}$

Displacement: Case 1 

Case 2:  

If the object is travelling from -3 to -1, then displacement 

If the object is travelling from $-3$ to $-1$, then displacement

$=x_{f}-x_{i}$

Displacement: Case 2 

Case 3:  

If the object is travelling from +4 to +2, then displacement 

Displacement: Case 3

Case 4:  

If the object follows the path depicted in the figure, the final and initial positions are the same, implying that the displacement is zero.

Displacement: Case 4 

We can conclude from the preceding examples that a body's displacement is positive if its final position is on the right side of its initial position and negative if its final position is on the left side of its initial position. The displacement of a moving object is said to be zero when it returns to its original position. Consider an athlete running in a clockwise direction along a circular track with radius r, beginning at A.

A Circular Track of Radius r 

What is the athlete's total distance traveled when he arrives at point B?

The athlete's total distance traveled when he arrives at point B equals to half of the circumference of the circular track, that is, $\frac{2 \pi r}{2}=\pi r$.

Displacement $=A B=2 r=$ Diameter of circle (the shortest distance between the initial and final positions).

If the athlete arrives at the starting point $A$, the distance covered is equal to the circumference of the circular track, i.e., $2 \pi r .$ However, the displacement is zero because the athlete's initial and final positions are the same.

Difference between Distance and Displacement

Motion: 

Uniform Motion and Non-uniform Motion: 

The distances covered by car A and car B with respect to time is given below:

Car A: 

Car B: 

The car A travels equal distances in equal time intervals, whereas the car B does not travel equal distances in equal time intervals. That is, car A's motion is an example of uniform motion, whereas car B's motion is an example of non-uniform motion.

A body is said to describe uniform motion when it covers equal distances in equal intervals of time.

When a body moves unequal distances in equal time intervals, or vice versa, this is referred to as non-uniform motion.

Speed: 

Ram and Krishna compete in various races over varying distances. Ram covers $1000 \mathrm{~m}$ in 20 minutes and Krishna covers $700 \mathrm{~m}$ in 10 minutes. Who is the fastest?

To determine who is faster, we will calculate the distance they cover in one minute.

Distance covered by Ram in one minute $=\frac{1000 \mathrm{~m}}{20 \mathrm{~min}}=500 \mathrm{~m} / \mathrm{min}$

Distance covered by Krishna in one minute $=\frac{700 \mathrm{~m}}{10 \mathrm{~min}}=70 \mathrm{~m} / \mathrm{min}$

Krishna covered more ground in the same amount of time. We conclude that Krishna is the faster of the two. 

Speed is defined as the distance traveled by a moving object in one unit of time.

$\text { speed }=\frac{\text { distance }}{\text { time }}=\frac{\mathrm{S}}{\mathrm{t}}$

Where S denotes the distance traveled and t denotes the time spent.

The SI unit of speed is millimeters per second (m/s). Speed is defined as a scalar quantity.

Uniform Speed:

The graph depicts the distance traveled by a ball every 2 seconds.

Every 2 seconds, the ball travels 10 meters. At any point between A and E, the ball moves at a speed of 5 m/s. The object is moving at a constant speed.

If an object travels the same distance in the same amount of time, it is said to be moving at a uniform speed.

Surface friction or resistance is ignored in this case.

Variable Speed or Non-Uniform Speed: 

The distance covered varies with time.

Variable Speed:  

For example, when a rubber ball is dropped from a certain height (h 1 ), it bounces up to a height less than the initial one (h 2 ). It keeps bouncing, but the height to which it rises keeps decreasing (h 3 , h 4 ). The ball's distance traveled per unit time decreases. The ball's speed varies from point to point. This type of speed is known as variable speed.

Average Speed and Instantaneous Speed:

When we travel by car, the speed varies depending on the road conditions at the time. The speed is calculated in this case by dividing the total distance traveled by the vehicle by the total time required for the journey. This is known as the average speed. 

The average speed of an object traveling S 1 in time t 1 , S 2 in time t 2 , and S n in time t n is given by,

$\text { Average speed }=\frac{\mathrm{S}_{1}+\mathrm{S}_{2}+\mathrm{S}_{3}+\ldots+\mathrm{S}_{\mathrm{n}}}{\mathrm{t}_{1}+\mathrm{t}_{2}+\mathrm{t}_{3}+\ldots+\mathrm{t}_{\mathrm{n}}}$

When we say that the car travels at an average speed of 60 km/h, we do not mean that it will travel at that speed for the duration of the journey. The actual speed of the car may be less than or greater than the average speed at a given location.

The speed of a moving body at any given point in time is referred to as instantaneous speed.

The diagram below depicts the various routes Shyam can take from his house to school. 

Shyam drives himself to school every day, averaging 60 km/h. Is it possible to find out how long it will take to get to the destination? Yes, you can use the relation to determine the time.

$\text { speed }=\frac{\text { distance }}{\text { time }}$

But you don't know what path he would have taken. As a result, simply providing the speed of a moving object does not allow one to determine the exact position of the object at any given time. As a result, there is a need to define a quantity that has both magnitude and direction.

Starting with A, consider two objects P and Q. Allow them to travel equal distances in equal time intervals, i.e. at the same speed. Can you guess where each of them will be in 20 seconds? P and Q are free to move in any direction. To determine the exact position of P and Q, we must also know their direction of motion.

Pictorial Representation of the Position of the Objects P and Q:

As a result, another physical quantity known as velocity is introduced to provide us with an idea of both speed and direction.

Velocity is defined as the distance traveled in a given direction by a moving object in a given time or speed in a given direction.

$\operatorname{velocity}(v)=\frac{\text { distance travelled in a specified direction }(s)}{\text { time taken }(t)}$

Note: 

Velocity is defined as the distance traveled in a given direction in a given amount of time. Displacement is the distance traveled in a specified direction.

As a result, velocity can be defined as the rate at which displacement changes.

Uniform Velocity and Non-Uniform Velocity:

Assume that two athletes, Ram and Shyam, are running at a constant speed of 5 m/s. Ram moves in a straight line, while Shyam follows a circular path. For a layperson, both Ram and Shyam are moving with uniform velocity, but for a physicist, only Ram is running with uniform velocity because his speed and direction of motion do not change.

In the case of Shyam, who is running on a circular track, the direction of motion changes at every instant because a circle is a polygon with infinite sides, and Shyam must change his direction at every instant.

A body is said to be moving with uniform velocity if it travels the same distance in the same amount of time in the same direction.

A body is said to be moving with variable velocity if it covers unequal distances in equal intervals of time and vice versa in a specified direction, or if its direction of motion changes.

Acceleration:

We are all aware that a car moving down the road does not have a uniform velocity. Either the speed or the direction of travel shifts. We say that a vehicle is accelerating when it is speeding up, i.e. when the speed increases.

Let us look at the change in velocity of a train traveling from Bangalore to Mysore to get an idea of acceleration. The train, which was initially at rest, begins to move; its velocity gradually increases until it reaches a constant velocity after a certain time interval. As the train approaches the next station, its speed gradually decreases until it comes to a halt.

When a train starts from a stop, its speed increases from zero, and we say it is accelerating. After a while, the speed becomes uniform, and we say that the train is moving at a uniform speed, which means that it is not accelerating. However, as the train approaches Mysore, it slows down, indicating that the train is accelerating in the opposite direction. When the train comes to a halt in Mysore, it stops accelerating once more.

As a result, it is clear that the term "acceleration" does not always imply that the speed of a moving body increases; it can also decrease, remain constant, or become zero.

In general, acceleration is defined as the rate at which the velocity of a moving body changes over time.

This change could be a change in the object's speed, direction of motion, or both.

Let us now look up a mathematical formula for calculating acceleration.

If an object moves with an initial velocity 'u' and reaches a final velocity 'v' in time 't,' then the acceleration 'a' produced by the object is

Acceleration = Rate at which velocity changes over time.

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

Unit of Acceleration:

The SI unit of acceleration is m/s 2 and it is a vector quantity.

Different Types of Acceleration:  

It is clear from the preceding example that acceleration takes various forms depending on the change in velocity.

Positive acceleration:

When an object's velocity increases, it is said to be moving with positive acceleration.

Positive Acceleration 

Example: a ball rolling downhill on an inclined plane.

Negative acceleration:

When an object's velocity decreases, it is said to be moving with negative acceleration. Negative acceleration can also be referred to as retardation or deceleration.

(1) A ball moving up an inclined plane.

(2) A vertically thrown upwards ball has a negative acceleration as its velocity decreases over time.

Zero Acceleration:  

If the change in velocity is zero, indicating that the object is either at rest or moving at uniform velocity, the object is said to have zero acceleration.

A parked car, for example, or a train moving at a constant speed of 90 km/hr.

Uniform Acceleration:  

The object is said to be moving with uniform acceleration if the change in velocity at equal intervals of time is always the same.

As an example, consider a body falling from a great height towards the earth's surface.

Non-uniform or Variable Acceleration:  

If the change in velocity over equal time intervals is not the same, the object is said to be moving with variable acceleration.

Motion:  

Distance-Time Table and Distance-Time Graph: 

Mr. X is taking a bus from New Delhi to Agra and recording his observations.

According to the table above, the bus travels equal distances at equal times. The bus is moving at a constant speed. In such a case, we can compute the distance traveled by the bus at any given point in time.

Consider an object moving from its initial position x i to its final position x f in time t at a uniform speed v. 

$\text { uniform speed }=\frac{\text { total distance }}{\text { time taken }}$

$v=\frac{x_{f}-x_{i}}{t}$

$x_{f}-x_{i}=v t \cdots \cdots(1)$

The relationship between distance, time, and average speed is given by equation (1). This relationship can be used to generate distance-time tables as well as to determine the position of any moving object at any given time. However, it is a time-consuming and tedious process, especially when we need to determine the position after a long period of time or compare the motion of two objects. In such cases, graphs such as the distance-time graph can be useful. A distance-time graph is a line graph that shows how distance changes over time. A distance-time graph plots time along the x-axis and distance along the y-axis. 

Distance-Time Graph for Non - Uniform Motion  

Let us now look at the nature of a distance-time graph for a non-uniform motion. The distance traveled by a bus every 15 minutes is shown in the table below.

We can deduce from the above table that the motion is non-uniform, i.e. it covers unequal distances in equal time intervals.

Measure time along the x-axis and distance along the y-axis.

Analyze the provided data and select the appropriate scale for time and distance.

Plot the points.

Join the points.

Consider any two points (A, B) on the graph.

Draw perpendicular from A to B to x and y axes.

Join A to C to get a right angled ACB.

The slope of the graph is shown below.

$\mathrm{AB} =\frac{\mathrm{BC}}{\mathrm{AC}}$

$=\frac{\mathrm{S}}{\mathrm{t}}$

$=\mathrm{speed}$

Write the title and scale chosen for the graph.

$\text { speed }=\frac{15-5}{30-15}$

$=\frac{10}{15}$

$=\frac{2}{3}$

$=0.666 \mathrm{~km} / \mathrm{min}$

Consider another two points on the graph, P and Q, and draw a right-angled triangle PRQ.

$\text { slope }=\text { speed }$

$\mathrm{PQ}=\frac{\mathrm{QR}}{\mathrm{PR}}$

$=\frac{35-30}{90-75}$

$=\frac{5}{15}$

$=0.333 \mathrm{Km} / \mathrm{min}$

Complete Graph

Nature of S- t Graph for Non- Uniform Motion and Uses of Graphs

Let us now see the nature of S-t graph for non-uniform motion. 

Nature of s-t Graph for Non-Uniform Motion: 

Figure (a) depicts the S-t graph as the speed of a moving object increases, while Figure (b) depicts the S-t graph as the speed of a moving object decreases. The nature of the S-t graph allows us to determine whether the object is moving at a constant or variable speed.

Uses of Graphical Representation:  

Because it provides a visual representation of two quantities, graphical representation is more informative than tables (e.g., distance vs. time)

A graph provides more information than a table at a glance. Both of the graphs shown here depict increasing speed.

Figure (1) depicts the nature of the variation in speed, indicating that the increase is greater in the beginning up to time t 1 and relatively lower after t 2 .

Similarly, fig (2) depicts how the increase in speed becomes greater after t1. A similar explanation applies to the decreasing speed.

Graphs are simple to read at a glance.

Graph plotting takes less time and is more convenient.

Graphs can be used to determine the position of any moving object at any point in time.

Two moving objects' motions can be easily compared.

Graphs reveal information about the nature of motion.

Velocity-Time Graph:  

The variation of velocity with time can be graphically represented to calculate acceleration in the same way that we calculated speed from the distance-time graph.

Let us now create a velocity-time(v-t) graph using the data below.

Draw time on the x-axis and velocity on the y-axis. 

Analyze the provided data and select the appropriate scale for the x and y axes.

Plot the Given Points.

Join the Points

Consider Any Two Points A and B on the Straight-Line Graph.

Draw Perpendiculars from A and B to x and y-axes.

Join A to C, ACB forms a Right-Angled Triangle.

Slope of the Graph

$\mathrm{AB}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\text { Change in velocity }}{\text { time }}=\text { acceleration }$

Calculations:

$\text { Acceleration }=\frac{30-20}{6-4}$

$=\frac{10}{2}$

$=5 \mathrm{~m} / \mathrm{s}^{2}$

Write the title for the graph.

V - T Graph:

Let us now examine the nature of the v - t graph for various types of motion.

a) Increasing Acceleration:

Uniform Acceleration

Non-Uniform Acceleration

(b) Decreasing Acceleration:

Non-uniform Retardation

Uniform Retardation

Zero Acceleration

Uses of Velocity-time Graphs  

The velocity-time graph can be used to derive the following results. 

The acceleration produced in a body. 

The distance traveled by a moving object. 

The equations of motion.

Speed - Time Graph

To compute the distance traveled by a moving object using a speed-time graph. 

The graph below depicts the speed-time graph of a car traveling at a constant speed of 60 km/h for 5 hours .

Speed-Time Graph of a Car Moving with Uniform Speed 

Distance travelled by the car,

Distance travelled by the car

$(S)=v \times t$

$=60 \times 5$

$=300 \mathrm{~km}$

But $60 \mathrm{~km} / \mathrm{h}=\mathrm{OC}=$ breadth of the rectangle $\mathrm{OABC}$

$5 h=O A=$ length of the rectangle $O A B C$

i.e., the distance covered by the car = length $\times$ breadth $=300 \mathrm{~km}$.

To calculate the distance traveled by a moving object using a speed-time graph, find the area enclosed by the speed-time graph and the time axis. In the case of non-uniform motion, the distance covered by the object increases in steps as the object's speed increases. During the time intervals $0-t_{1}, t_{1}-t_{2}, t_{2}-t_{3}, \ldots \ldots$, the speed remains constant.

The motion of an object moving at a variable speed is depicted in the figure below.

Speed - Time Graph for an Object Moving with Variable Speed

Calculation of Distance:

The object's total distance traveled during the time interval. 

0- t 6 = Area of rectangle 1 + area of rectangle 2 + …… + area of rectangle 6. 

Motion 

Equations of Motion  

Time, speed, distance covered, and acceleration are the variables in a uniformly accelerated rectilinear motion. These quantities have simple relationships. These relationships are expressed using equations known as equations of motion. 

The equations of motion are: 

(1) $v=u+a t$

(2) $S=u t+\dfrac{1}{2} a t^{2}$

(3) $v^{2}-u^{2}=2 a S$

Derivation of the First Equation of Motion

Consider a particle moving in a straight line with a constant acceleration ‘a’. Let the particle be at A at t=0, and u be its initial velocity, and v be its final velocity at t=t.

$a=\dfrac{v-u}{t}$ 

$v-u=a t$ 

I Equation of Motion

Second Equation of Motion

$\text { Average velocity } =\dfrac{\text { total distance travelled }}{\text { total time taken }}$

$=\dfrac{S}{t} \ldots \ldots(1)$

Average velocity can alsobe written as

$\frac{u+v}{2} \cdots \cdots(2)$

From equation (1) and (2),

$\frac{S}{t}=\frac{u+v}{2} \cdots \cdots(3)$

The first equation of motion is $v=u+a t$. Substituting the value of $v$ in equation (3), we get

$\frac{S}{t}=\frac{u+u+a t}{2}=\frac{(u+u+a t) t}{2}$

$\quad=\frac{(2 u+a t) t}{2}$

$S=u t+\frac{1}{2} a t^{2}$

II equation of motion

Third Equation of Motion

The first equation of motion is

$v-u=a t \ldots(1)$

Average velocity $=\frac{S}{t} \cdots \cdots(2)$

Average velocity $=\frac{u+v}{t} \cdots \cdot(3)$

From equation (2) and equation (3) we get,

$\frac{u+v}{t}=\frac{S}{t} \cdots \ldots(4)$

Multiplying equation (1) and equation (4) we get,

$(v-u)(v+u)=a t \times \frac{2 S}{t}$

$(v-u)(v+u)=2 a S$

$\left(v^{2}-u^{2}\right)=2 a S$

III equation of motion

Derivations of Equations of Motion (Graphically) 

First Equation of Motion

Consider an object moving in a straight line with a uniform velocity u. When its initial velocity is u, give it a uniform acceleration an at time t = 0. The object's velocity increases as a result of the acceleration to v (final velocity) in time t, and S is the distance covered by the object in time t.

The graph depicts the velocity-time graph of the object's motion.

The acceleration of a moving object is given by the slope of the v - t graph.

Thus, acceleration = slope

$\mathrm{AB}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{v-u}{t-0}$

I equation of motion

Second Equation of Motion  

Let u be an object's initial velocity and 'a' be the acceleration produced in the body. The area enclosed by the velocity-time graph for the time interval 0 to t gives the distance travelled S in time t.

Graphical Derivation of Second Equation

Distance travelled S = area of the trapezium ABDO

Distance travelled $S=$ area of the trapezium ABDO

$=$ area of reactangle $A C D O+$ area of $\triangle A B C$

$=\mathrm{OD} \times \mathrm{OA}+\frac{1}{2} \mathrm{BC} \times \mathrm{AC}$

$=\mathrm{t} \times \mathrm{u}+\frac{1}{2}(\mathrm{v}-\mathrm{u}) \times \mathrm{t}$

$=\mathrm{ut}+\frac{1}{2}(v-u) \times t$

$v=u+$ at $\mid$ equation of motion; $v-u=a t$

$S=u t+\frac{1}{2} a t \times t$

$=u t+\frac{1}{2} a t^{2}$

Third Equation of Motion  

Let 'u' be an object's initial velocity and a be the acceleration produced in the body. The area enclosed by the v - t graph gives the distance travelled 'S' in time 't'.

Graphical Derivation of Third Equation

$\mathrm{S}=$ area of trapezium $\mathrm{OABD}$

$=\frac{1}{2}\left(b_{1}+b_{2}\right) h$

$=\frac{1}{2}(\mathrm{OA}+\mathrm{BD}) \mathrm{AC}$

$=\frac{1}{2}(u+v) t \cdots \cdots(1)$

But we know that $a=\frac{v-u}{t}$ or $t=\frac{v-u}{a}$

Substituting the value of $t$ in equation (1) we get,

$S=\frac{1}{2} \frac{(u+v)(v-u)}{a}$

$=\frac{1}{2} \frac{(v+u)(v-u)}{a}$

$2 a s=(v+u)(v-u)$

$(v+u)(v-u)=2 a s$

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

III Equation of Motion 

Circular Motion 

We classified motion along a circular track as non-uniform motion in the example discussed under the topic uniform and non-uniform motion. Let's take a look at why circular motion is considered non-uniform motion. The figure depicts an athlete running at a constant speed on a hexagonal track.

Athlete Running on a Regular Hexagonal Track 

The athlete runs at a constant speed along the track segments AB, BC, CD, DE, EF, and FA, and at the turns, he quickly changes direction to stay on the track without changing his speed. Similarly, if the track had been a regular octagon, the athlete would have had to change directions eight times in order to stay on the track.

Athlete Running on a Regular Octagonal Track 

The athlete must turn more frequently as the number of sides of the track increases. If we increase the number of sides indefinitely, the track will take on the shape of a circle. As a result, because a circle is a polygon with infinite sides, motion along a circular path is classified as non-uniform motion.

Athlete Running on a Circular Track 

Thus, an object moving at uniform speed along a circular track is an example of non-uniform motion because the object's direction of motion changes at every instant of time. 

Examples of Uniform Circular Motion 

(1) A car negotiating a curve at a constant speed.

(2) An athlete spinning a hammer in a circle before throwing it.

(3) An aircraft looping the loop.

Expression for Linear Velocity

If an athlete takes t seconds to complete one circular path of radius r, then the velocity v is given by the relation,

$\mathrm{v}=\frac{\text { distance travelled }}{\text { time }}$

Distance travelled = circumference of the circle

$=2\pi\mathrm{r}$

Linear velocity $=\frac{\text { 2r }}{\mathrm{t}}$

Class 9 Science Chapter 8 Revision Notes - Free PDF Download

Contents of class 9 science motion notes.

The Class 9 science notes chapter 8 contains detailed discussions on the topic of motion. It tells you about the basic definition of motion and rest. Then the chapter discusses the different types of motion. Next comes distance and displacement. 

With all this knowledge in your hand, we thereby venture to find out the speed and velocities of different objects. We also look into acceleration and its types. 

The notes also contain worked-out examples of numerical problems and other questions to give us a clear idea of the type of questions we can expect in the examination.

Subtopics Covered in Class 9 Chapter 8 Science Notes

The subtopics covered in class 9 science ch 8 notes are:

Motion and Rest: The change in position of an object with time and with respect to a stationary object is moving. If a body undergoes no change in position with time and with respect to a stationary object then it is said to be at rest

Types of Motion: Motion is divided into the following types:

Translation Motion: Motion along a straight line or curved path

Vibrational Motion: Motion of a particle to and from fixed positions.

Rotatory Motion: The circular motion of an object around the axis of rotation.

Distance and Displacement: Distance is defined as the total length of the path traveled by an object. Displacement on the other hand is the net effective change in position. In this section, we shall also look into various examples of numerical problems that aim at finding the distance and displacement of various objects.

Uniform and Non-Uniform Motion: Motion is said to be uniform when the rate of change of position remains the same across intervals multiple times with respect to the same object. Non-uniform motion sees different rates of change of position with respect to the object of reference.

Speed: Uniform And Variable: Speed is defined as the rate of change of position. It can either be uniform or variable.

Velocity: The velocity of an object is defined as the rate of change of displacement.

Acceleration: Acceleration is the rate of change of velocity with respect to time.

Basics of Plotting a Graph: Graphs are basically illustrations that are used to show the rate of change of a character. A graph is made by the intersection of two mutually perpendicular lines. Each of these lines is called the axis. 

How to work out time vs distance or displacement

Laws of Motion: Motion is governed by laws of physics known as Newton’s law of motion or simply the laws of motion. 

Ch 8 science class 9 notes are very important from the point of view of examinations as the chapter of motion has a net weightage of 9. Apart from that this chapter is especially important for candidates who want to pursue science in higher classes.

Why Choose Notes of ch 8 Science Class 9 by Vedantu?

Vedantu's class 9th science chapter 8 notes are the greatest study companion available. Physics may be a difficult topic to master, especially on your own. Vedantu's Motion class 9 notes are designed just for students. Despite the fact that it is developed by specialists, the language used is relatively easy. The subject of motion is both wide and vital. It is critical for students to understand as much as they can about motion without becoming overburdened. This is why Vedantu's notes give the ideal blend of study material that is both easy and dense with knowledge. Some advantages of utilising Vedantu include:

It is easy to access

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It is free to download from the official website of Vedantu

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The chapter 'Motion' may provide pupils with a wealth of information. All of the major topics are discussed here. The students are required to understand the important coverage so that they can prepare accordingly, this can be done by studying these notes from the chapter 'Motion'.

The students can also download the free pdf which is mentioned in this article. 

FAQs on Motion Class 9 Notes CBSE Science Chapter 8 (Free PDF Download)

1. When can an object have a zero displacement according to what you have studied in Notes of the Chapter Motion?

If an item goes along a certain path and then returns to its original position, the associated displacement of that object is zero. The length of the path will be travelled, but because the beginning and ending points of the movement are the same, the displacement will be zero.

2. Explain the type of motions according to what you have studied in Notes of the Chapter Motion.

Motion is the term given to the change in position of an object with respect to time. There are three types of motions:

Rotatory Motion: This takes place in a circular path when the object moves around the axis of rotation.

Vibrational Motion: This occurs when the object moves to and from a fixed position.

Translation Motion: This is caused by an object moving along a curved path or a straight line.

3. What are the subtopics covered in Class 9 Science Chapter 8 notes?

The CBSE Class 9 Science Chapter 8 - Motion Revision Notes provide well-explained details on each topic that has been covered in the NCERT and is a part of the latest syllabus provided by CBSE for Class 9 Science. These subtopics include the following:

Motion and Rest

Types of Motion

Distance and Displacement

Uniform and Non-uniform Motion

Acceleration

Basics of Plotting a Graph

Newton’s Laws of Motion

4. What are the benefits of referring to Vedantu’s Class 9 Science Chapter 8 notes?

Referring to CBSE Class 9 Science Chapter 8 - Motion Revision Notes by Vedantu can provide students with several benefits like:

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Save time by providing notes that are ready for studying

Help in increasing clarity about the different concepts taught in chapter 8

Help in understanding and solving numerical that are an important part of this chapter

Free of cost PDF download for studying offline

5. What are the Laws of Motion discussed in Chapter 8 Class 9 Science?

Newton established three laws of motion. According to the first law of motion, an object at rest or in uniform motion will remain in that condition until an external force is applied to it. Newton's second law states that the acceleration of an item is proportional to the force applied to it and its mass. According to the third and final law, every action has an equal and opposite response. Students may learn more by visiting the Vedantu website or app.

STUDY MATERIALS FOR CLASS 9

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