Motion - class 9 Notes, formulae, and Numericals

Motion – Class 9 Notes, Formulae, and Numerical PDF Download

Chapter: Motion

Class: 9 Science

Target: CBSE Board Exams | State Board Exams | NTSE | Olympiads | Foundation 

Introduction to Motion

As we know our world is always moving — cars speeding, fans spinning, water falling. The scientific study of how things move is called Motion. This chapter helps us understand the types, causes, and equations of motion with real-life examples and graphical analysis.

Important Keywords:

• Motion concept and definition

• Types of motion with Examples

• Reference Point

• Distance
  

• Displacement

• Speed
  

• Velocity
  

• Acceleration

• Graphical Representation of Motion

• Equation of Motion by Graphical method

• Uniform Circular Motion (UCM)

• CBSE Class 9 Physics Numerical

Motion concept and Definition

An object is said to be in motion if it changes its position with time relative to a reference point.

Examples:

                                - A moving train

                                - A rotating fan

                                - A falling apple

If the object doesn’t change its position with respect to time, it is said to be at rest.

Types of Motion with Examples

1. Translation Motion

- The whole body moves from one point to another.

- Can be Rectilinear (in a straight line) or Curvilinear (along a curved path)

Example: Car moving on a straight road

2. Rotational Motion

- The object rotates around a fixed axis.

Example: Earth’s rotation, spinning top

3. Oscillatory Motion

- Repeated back and forth movement about a mean position.

Example: Swing, pendulum

 Reference Point

Definition: A reference point is a stationary object or location that is used to compare and judge the motion of another object.

 Example: If you’re sitting in a moving train and look outside, the trees seem to move backward — because you are using the trees (which are stationary) as your reference point. But if you look at a fellow passenger, they appear still — because they are moving with you, relative to you.

1. Distance 

Definition:

Distance is the total path length traveled by an object during its motion, irrespective of direction.

- It is a scalar quantity (only magnitude, no direction)

- Always positive or zero.

- Represented by 'd' or sometimes 's' in basic equations.

Example:

If you walk 7 m forward and then 5 m backward,

Total distance = 7 + 5 = 12 m

for more about Distance click here

2. Displacement

Definition:

Displacement is the shortest straight line distance between the initial and final position of the object, with direction.

- It is a vector quantity (magnitude + direction)

- It can be positive, negative or zero

- It is represented by 's'

Example:

In the same case:

If you walk 7 m forward and then 5 m backward,

Total displacement = 7 - 5 = 2 m

for more about Displacement click here

3. Speed 

Definition:

- Speed is the rate at which an object covers distance.

- It tells us how fast something is moving, irrespective of direction.

- It is a scalar quantity. 

For more about Speed click here

Formula:

                                          Speed = Distance/time

Units

- SI unit m/s (meter per second)

- Other units km/hr, cm/s

- 1 km/hr = 5/18 m/s 

Average Speed

When an object travels different distances in different time intervals, we calculate the overall or total speed using Average Speed.

For more about Average Speed click here

Average Speed Formula: 

                                    Average Speed = Total Time Taken / Total Distance Traveled

- It does not depend on how the speed changes during the journey — only the total distance and total time matter.

Units

- SI unit m/s (meter per second)

- Other units km/hr, cm/s

- 1 km/hr = 5/18 m/s

Type 1: Two Speeds, Equal Distance

Q1. A car travels a certain distance at a speed of 40 km/h and returns the same distance at a speed of 60 km/h. What is its average speed for the entire journey?

Solution:

Let distance = d
Time while going = d/40
Time while returning = d/60
Total time = d/40 + d/60 = (5d + 3d)/120 = 8d/120 = d/15
Total distance = 2d
Average speed = Total distance / Total time = 2d / (d/15) = 30 km/h

Type 2: Two Speeds, Unequal Distance

 Q2. A cyclist covers 20 km at 10 km/h and then 30 km at 15 km/h. Find the average speed of the entire journey.

 Solution:

Time for 1st part = 20/10 = 2 h
Time for 2nd part = 30/15 = 2 h
Total distance = 20 + 30 = 50 km
Total time = 2 + 2 = 4 h
Average speed = 50/4 = 12.5 km/h

Type 3: Time-based Journey

Q3. A student walks to school for 30 minutes at a speed of 4 km/h, then returns home by car in 10 minutes at a speed of 30 km/h. What is the average speed for the entire trip?

Solution:

Walking distance = 4 × (30/60) = 2 km
Car distance = 30 × (10/60) = 5 km
Total distance = 2 + 5 = 7 km
Total time = 30 + 10 = 40 min = 40/60 = 2/3 h
Average speed = 7 ÷ (2/3) = 10.5 km/h

Type 4: Variable Speeds and Time

Q4. A train runs for 2 hours at 60 km/h, then for 1.5 hours at 80 km/h. What is its average speed?

 Solution:

Distance1 = 60 × 2 = 120 km
Distance2 = 80 × 1.5 = 120 km
Total distance = 240 km
Total time = 2 + 1.5 = 3.5 h
Average speed = 240 / 3.5 = 68.57 km/h

Type 5: Using m/s and converting units

Q5. A person walks 300 m at 1.5 m/s and then runs 600 m at 3 m/s. Calculate his average speed in m/s and km/h.

Solution:

Time1 = 300 / 1.5 = 200 s
Time2 = 600 / 3 = 200 s
Total distance = 300 + 600 = 900 m
Total time = 400 s
Average speed = 900 / 400 = 2.25 m/s
Convert to km/h: 2.25 × 18/5 = 8.1 km/h

For more numerical on speed and average speed click here

4. Velocity

Definition:

- Velocity is the rate of change of displacement of an object with respect to time. 
- It is a vector quantity, which means it has both magnitude and direction.

for more about Distance click here

Formula: 
                     Velocity = Displacement / Time

Units: 

- SI Unit: metre per second (m/s)
- Other units: kilometre per hour (km/h), cm/s
- Conversion: 1 km/h = 5/18 m/s

Type 1: Constant Velocity

Q1. A car moves 100 meters north in 10 seconds. What is its velocity?

Solution:

Velocity = Displacement / Time = 100 m / 10 s = 10 m/s (North)

Type 2: Negative Velocity

 Q2. A ball rolls 20 meters east and then 30 meters west in 10 seconds. What is the velocity of the ball?

Solution:

Displacement = 30 - 20 = 10 m (West)
Velocity = 10 m / 10 s = 1 m/s (West)

Type 3: Zero Velocity

 Q3. A student walks 50 meters to the east and returns to the starting point in 5 minutes. What is his velocity?

Solution:

Displacement = 0 (since he returned to start)
Velocity = 0 / 300 s = 0 m/s

Type 4: Velocity with Conversion

Q4. A train travels 36 km in 30 minutes. Find its velocity in m/s.

Solution:

Convert 36 km = 36,000 m and 30 min = 1800 s
Velocity = 36000 / 1800 = 20 m/s

Type 5: Direction-based Velocity

Q5. A boy runs 100 m south in 20 seconds, then 60 m north in 10 seconds. What is his average velocity?

Solution:

Net displacement = 100 - 60 = 40 m (South)
Total time = 30 s
Average velocity = 40 / 30 = 1.33 m/s (South)

5. Acceleration

Definition:

-          - Acceleration is the rate of change of velocity of an object with respect to time.

-          - It is a vector quantity.

 Formula:

Acceleration (a) = Final velocity (v) - Initial velocity (u) / Time (t) = v - u / t

Units:

-          - SI unit: meter per second square (m/s²)

-          -  CGS unit: centimeter per second square (cm/s²)

 Types of Acceleration:

1.        1. Uniform Acceleration:

·         Velocity changes equally in equal intervals of time.

·         Example: A freely falling body.

2.       2. Non-uniform Acceleration:

·         Velocity changes unequally in equal time intervals.

·         Example:  A car moving in traffic.

3.       3. Positive Acceleration:

·         Velocity increases with time.

·         Example: A speeding car.

4.       4. Negative Acceleration (Retardation):

·         Velocity decreases with time.

·         Example: A braking bicycle.

Graphical Representation of Motion.

Here is a detailed explanation of Displacement-Time (s-t) graph and Velocity-Time (v-t) graph with diagrams.

1. Displacement-Time (s-t) Graph

- It shows how displacement of an object changes with time.

- It gives velocity = slope of graph.

Cases:

1) Uniform motion (Constant velocity)

- Graph = Straight line

- Slope = Constant (Object moving with uniform speed)

2) Non-uniform motion

- Graph = Curve

- Slope = Changing

2. Velocity-Time (v-t) Graph

- It shows how velocity changes with time

- Area under graph = displacement

- Slope = acceleration

Cases:

1) Uniform motion (Constant velocity)

- Graph = Straight horizontal line

- Acceleration = zero

   

2) Uniform acceleration 

- Velocity increases at constant rate.

- Graph = slanting straight line.

3) Non-uniform acceleration

- Velocity increases or decreases regularly.

- Graph = Curve

4) Retardation (Uniform deceleration)

- It shows that velocity decreases uniformly over time.

- Graph = Straight line sloping downwards.

Equation of motion by Graphical method

We derive three important equations of motion for a body moving with uniform acceleration using the velocity-time graph:

Let:

                        Initial velocity =      u

                        Final velocity =       v

                        Time taken =            t

                        Acceleration =         a

                        Distance traveled =  s

1. First Equation of Motion: v = u + at

From the graph:

Slope = BC/AC

BC = change in velocity = v - u

AC = time = t

As,

slope of v-t graph gives acceleration.

Acceleration = Change in velocity / Time

a = (v - u) / t

=> at = v - u

=> v = u + at

This is the first equation of motion.

2. Second Equation of Motion: s = ut + ½ at²

Distance travelled = Area under velocity-time graph (OABD)

Area = Area of rectangle OACD + Area of triangle ABC

= (u × t) + ½ (v - u) × t

Using v = u + at => v - u = at

s = ut + ½ at × t = ut + ½ at²

This is the second equation of motion.

3. Third Equation of Motion: v² = u² + 2as

Distance = Area under velocity-time graph = Area of trapezium OABD

s = ½ (u + v) × t

Using t = (v - u) / a

s = ½ (u + v) × (v - u) / a

=> 2as = (v + u)(v - u) = v² - u²

This is the third equation of motion.

Uniform Circular Motion (UCM)

Definition:

- When an object moves in a circular path with constant speed, the motion is called Uniform Circular Motion (UCM)

- Even though the speed is constant, velocity is not, because the direction of motion continuously changes in a circular path.

Key Points:

•   Speed is constant, but velocity changes.
•  The motion is accelerated, because of continuous change in direction.
•   The acceleration is directed towards the center of the circle. This is called centripetal    acceleration.
•   A centripetal force is required to keep the object moving in a circle.
•   The object does not fly away because this force pulls it inward.

-         Formulae:

• Speed in circular motion:

                                                        Speed = 2πr / T

                                            Where,

                                                r = radius of circular path

                                                T = time period to complete one circle

• Centripetal Force:

                                                           F = mv² / r

                                               Where,

                                                            m = mass of object

                                                             v = speed

                                                             r = radius of the circle

Examples of Uniform Circular Motion:
1)    Motion of moon around Earth
2)    Artificial satellite orbiting the Earth
3)    Electron revolving around the nucleus
4)    Vehicle turning around a circular track with uniform speed.

Numerical:

1) A train accelerates from 36 km/h to 54 km/h in 10 sec.
        (i) Acceleration
        (ii) The distance traveled by the car.

Solution:-

Given:
Initial velocity u = 36 km/h = 10 m/s
Final velocity v = 54 km/h = 15 m/s
Time t = 10 s

(i) Acceleration (a):
v = u + at
15 = 10 + 10a
=> a = 0.5 m/s²

(ii) Distance travelled (s):
s = ut + (1/2)at² = 10×10 + 0.5×0.5×100 = 100 + 25 = 125 m

2) A truck travelling at 54 km/h is slowed down to 36 km/h in 10 sec. Find the retardation.

Solution:-

Given:

Initial velocity u = 54 km/h = 15 m/s
Final velocity v = 36 km/h = 10 m/s
Time t = 10 s

Using v = u + at:
10 = 15 + a×10
=> a = (10 - 15)/10 = -0.5 m/s² (retardation)

3) A car starts from rest and acquires a velocity of 54 km/h in 2 sec. Find
        (i) The acceleration
        (ii) Distance travelled by the car (assume motion is uniform)

Solution:-

Given:

u = 0, v = 54 km/h = 15 m/s, t = 2 s

(i) a = (v - u)/t = (15 - 0)/2 = 7.5 m/s²
(ii) s = ut + (1/2)at² = 0 + 0.5×7.5×4 = 15 m

4) A ball is thrown upwards and it goes to the height 100 m and comes down.
        (i) What is the net displacement?
        (ii) What is the net distance?

Solution:-

Given:

(i) Net displacement = 0 m (it comes back to the same point)
(ii) Net distance = 100 m (up) + 100 m (down) = 200 m

5) An object travels 20 m in 2 s and then another 16 m in 2 s. What is the average speed of the object?

Solution:-

Given:

Total distance = 20 + 16 = 36 m
Total time = 2 + 2 = 4 s
Average speed = Total distance / Total time = 36 / 4 = 9 m/s

6) A particle with a velocity of 2 m/s at t = 0 moves along a straight line with a constant acceleration of 0.2 m/s². Find the displacement of the particle in 10 s.

Solution:-

Given:

u = 2 m/s, a = 0.2 m/s², t = 10 s

s = ut + (1/2)at² = 2×10 + 0.5×0.2×100 = 20 + 10 = 30 m

Read also: Light-Reflection and Refraction Part-1 class 10th

Read also: Class 11th Chapter-1: Physical World Notes

Read also: How to study Physics for 11th and 12th

Also download: Practice Numerical on Motion worksheet download

Read also: Online Education: A way of Future Education

Practice Numerical:

1) A taxi driver noted the reading on the odometer fitted in the vehicle as 1052 km when he started the journey. After 30 minutes drive, he noted that the odometer reading was 1088 km. find the average speed of the taxi. (click here for Answer)

2) A boy is running on a straight road. He runs 500 m towards north in 2 minutes 10 seconds and then turns back and runs 200 m in 1 minute. Calculate

            i) His average speed and magnitude of average velocity during first 2 minutes 10 seconds and

            ii) His average speed and magnitude of average velocity during the whole journey. (click here for Answer)

 3) A train travels 20 km at a uniform speed of 60 km/h and the next 20 km at a uniform speed of 80 km/hr. calculate its average speed. (click here for Answer)

4) An object dropped from a leaf false with a constant acceleration of 10 m/s². Find its Speed 2 second after it was dropped. (Click here for Answer)

5) A body starts from rest and moves with a uniform acceleration of 2 m/s². What will be its velocity and displacement at the end of 10 seconds? (Answer:- 100 m)

6) A body moves with an initial velocity of 2 m/s and uniform acceleration of 3 m/s². Calculate the velocity when it has traveled a distance of 77 m. (Answer:- 21.6 m/s)

 7) An athlete completes one round of circular track of diameter 50 m in 10 second. What will be the distance cover and the displacement at the end of 45 second? (Answer:- 706.5 m and 50 m)

8) An object moves with an initial velocity of 10 m/s and uniform acceleration of 0.5 m/s². Calculate the velocity after 10 second and distance traveled in this time. (Answer:- 125 m )

9) An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 second if both objects drop with same acceleration, how does the difference in heights very with time? (Answer:- Difference = 50 m and difference remains constant with time )

10)  An object starting from rest travels 20 m in first 2 second and 160 m in next 4 second. What will be the velocity after 7 second from the start? (Answer:- 70 m/s)

Read also: Force and Laws of Motion class 9

Read also: Gravitation class 9 Notes, Practice Numerical

Read also: Online Education: A way of Future Education

Also download: Practice Numerical on Motion worksheet download

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