Motion - class 9 Notes, formulae, and Numericals





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

Chapter: Motion

Class: 9 Science

Target: CBSE 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 Numericals

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

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

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.
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.

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

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.

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

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