Science! Physics Notes on the topic - 'MOTION'. For Class:- 9th {with ncert examples} Created By - 'Neha Rohtagi'. I hope that you will found this presentation useful and it will help you out for your concept understanding. Thank You! Please give feedbacks and suggestions to get presentations on more interesting topics.

1. Motion
A body is said to be in Motion; when it's position changes
continuously with reference to a point.
Motion can be of different types depending upon the
path followed by the object:
Linear Motion - Object moves in a straight line.
1.
Circular Motion - Object moves in a circular path.
2.
Oscillatory Motion - Object shows to & fro
movement.
3.
Scalar Quantity: A physical quantity which depends
on magnitude but not on direction. For e.g. Distance,
Speed, etc.
Vector Quantity: A physical quantity which depends
both on magnitude and direction. For e.g.
Displacement, Velocity, etc.

2. Distance and Displacement:
Distance: The actual path travelled by the body from it’s
initial position to the final position.
It is a scalar quantity.
Displacement: It is the shortest path covered by an object to
reach the final position.
It is a vector quantity, can be Zero or Negative.
Distance Displacement

3. Uniform And Non - Uniform Motion:
Uniform Motion: When a body travels equal distance in
equal intervals of time, then it is said to be Uniform Motion.
Non - Uniform Motion: When a body travels unequal
distance in equal intervals of time, then it is said to be Non -
Uniform Motion.
Two types of Non - Uniform Motion Are:
Accelerated Motion: When motion of body increases with
time.
De - Accelerated Motion: When motion of body decreases
with time.

4. Speed:
The measurement of distance travelled by a body per unit
time is called Speed.
The SI Unit of speed is m/s.
It is a Scalar Quantity.
Speed = Distance travelled/ Time taken by the body.
In Uniform Motion, the speed of body remains constant.
In Non - Uniform Motion, the speed of the body will not
remain constant and vary with time period.
Average Speed = Total Distance travelled/ Total time taken.
Examples 8.1
An object travels 16m in 4s and then another 16m in 2s. What
is the average speed of the object?
Ans. Total distance travelled by the object = 16m + 16m = 32m.
Total time taken = 4s + 2s = 6s.
Avg. Speed = 32m/ 6s = 5.33 m/s.
CONVERSION FACTOR:
Change km/hr to m/s- 1000m/ (60) (60) = 5/18 m/s.

5. Velocity:
The measurement of distance travelled by a body per unit
time in a particular direction is called Velocity.
The SI Unit of velocity is m/s.
Velocity = Displacement/ Time taken by the body.
It is a Vector Quantity.
Average Velocity = Total Displacement/ Total time.
It can be positive, negative or zero.
Example 8.2
The odometer of a car reads 2000 km at the start of a trip and
2400 km at the end of the trip. If the trip took 8 h, calculate
the average speed of the car in km/h and m/so.
Ans. Distance covered = 2400 km – 2000 km = 400 km.
Time elapsed, t = 8 h
Average speed = 400km/ 8h = 5km/h = 13.9 m/s.
Example 8.3
Usha swims in a 90 m long pool. She covers 180 m in one
minute by swimming from one end to the other and back
along the same straight path. Find the average speed and
average velocity of Usha.
Ans. Total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in 1 min = 0 m
Avg. Speed= 180m/ 60s = 3m/s.
Avg. Velocity= 0m/ 60s= 0m/s.

6. Acceleration:
It refers to the rate of change in velocity and is generally
observed in Non- Uniform Motion.
The SI Unit of velocity is m/s^2.
Acceleration = Change in Velocity/ Time taken by the body.
= v - u/t
Where v = final velocity, u = initial velocity and t = time
taken.
It is a Vector Quantity.
It can be positive, negative or zero.
Retardation:
Also known as ‘Deceleration’.
Retardation is defined as the rate of decrease of velocity
with time.
Generally referred to as negative of acceleration and
represented using a Negative Sign.
Ex
ample 8.4
Starting from a stationary position, Rahul paddles his bicycle
to attain a velocity of 6 m s–1 in 30 s. Then he applies brakes
such that the velocity of the bicycle comes down to 4 m s-1 in
the next 5 s. Calculate the acceleration of the bicycle in both
the cases.

7. Ans. In the first case:
initial velocity, u = 0
final velocity, v = 6m/s
time, t = 30s
Therefore, acceleration, a = v - u/t
Substituting the given values of u,v and t in the above equation,
we get: 6m/s - 0/ 30s = 0.2m/s^2.
In the second case:
initial velocity, u = 6m/s
final velocity, v = 4m/s
time, t = 5s.
Therefore, acceleration, a = v - u/t
Substituting the given values of u,v and t in the above equation,
we get: 4m/s - 6m/s / 5s = -0.4m/s^2.
Graphical representation Of Motion:
Distance - Time Graph: Uniform Speed
1.
The change in the position of an object with time can be
represented on the Distance-Time Graph.
In this graph, time is taken along the x–axis and distance is
taken along the y-axis.

8. 2. Distance - Time Graph: Non - Uniform Speed
The nature of this graph shows non linear variation of the
distance travelled by the body with time.
3. Velocity - Time Graph: Uniform Motion
The variation in velocity with time for an object moving in a
straight line can be represented by a velocity-time graph.
In this graph, time is represented along the x-axis and the
velocity is represented along the y-axis.
4. Velocity - Time Graph: Uniform Acceleration
The nature of the graph shows that velocity changes by
equal amounts in equal intervals of time.
Thus, the velocity-time graph is a straight line.

9. 5. Velocity - Time Graph: Non - Uniform Acceleration
In the case of non-uniformly accelerated motion, velocity-
time graphs can have any shape.
Equations of Motion:
v = u + at
1.
s = ut + 1/2 at^2
2.
v^2 - u^2 = 2as
3.
Example 8.5
A train starting from rest attains a velocity of 72 km/h in 5
min. Assuming that the acceleration is uniform, find
(i) the acceleration and
(ii) the distance travelled by the train for attaining this
velocity.

10. Ans. u = 0
v = 72 km/h = 20 m/s t = 5 min = 300s.
1. Therefore, acceleration, a = v - u/t
= 20m/s - 0m/s/300s
= 1/15 m/s^2
2. v^2 - u^2 = 2as
s = v^2/ 2a = (20 m/s)^2/ 2 (1/15 m/s)
= 3000 m.
Example 8.6
A car accelerates uniformly from 18 km/h to 36 km/h in 5s.
Calculate
(i) the acceleration and
(ii) the distance covered by the car in that time.
Ans. u = 18 km/h = 5 m/s
v = 36 km/h = 10 m/s t = 5 min = 300s.
1. Therefore, acceleration, a = v - u/t
= 10m/s - 5m/s/ 5s
= 1 m/s^2
2. v^2 - u^2 = 2as
s = v^2/ 2a = (10 m/s)^2/ 2 (1 m/s^2)
= 37.5 m.
Example 8.7
The brakes applied to a car produce an acceleration of 6 ms^2
in the opposite direction to the motion. If the car takes 2s to
stop after the application of brakes, calculate the distance it
travels during this time.

11. Uniform Circular Motion:
When an object moves in a circular path with uniform speed,
its motion is called Uniform Circular Motion.
v = 2r/t
On being released the stone moves along a straight line
Tangential to the Circular Path. This is because once the
stone is released, it continues to move along the direction it
has been moving at that instant.
This shows that the direction of motion changed at every
point when the stone was moving along the circular path.

12. Thank You!