This document defines and explains key concepts related to motion including: - Motion is defined as a change in position or place over time. It can be linear or circular. - Relative motion occurs between two objects moving in the same or opposite directions. - Distance is the total path length traveled, while displacement is the shortest distance between the start and end points. - Speed is how distance changes over time, while velocity includes both speed and direction. - Acceleration is the rate of change of velocity over time and can be positive, negative, or zero. - Graphs of position vs time and velocity vs time can be used to analyze an object's motion.

2. WHAT IS MOTION??
DEFINITION:
• CHANGE IN POSITION OF AN OBJECT
WITH RESPECT TO TIME.
• ANY MOVEMENT (LINEAR OR
CIRCULAR) OR CHANGE IN PLACE
• MOTION OCCURS WHEN AN OBJECT
CHANGES ITS POSITION.
RELATIVE MOTION:
• WHEN TWO OBJECTS ARE MOVING IN A
PLANE (EITHER IN SAME DIRECTION OR
OPPOSITE) EACH HAVE RELATIVE
MOTION WITH RESPECT TO SECOND.
E.G. A PERSON SITTING IN A TRAIN AND
WATCHING A TREE, IN THIS CASE TREE
IS STABLE BUT IS ASSUMED TO BE
MOVING BUT WITH RESPECT TO TRAIN

3. Figure Showing change in the position of
the bear.
Figure Showing an instance of relative
motion

4. DISTANCE VS. DISPLACEMENT
DISTANCE
• TOTAL LENGTH OF PATH COVERED BY
AN OBJECT.
• IT HAS ONLY MAGNITUDE WITHOUT
DIRECTION. (SCALAR QUANTITY)
• EITHER ZERO OR POSITIVE.
DISPLACEMENT
• SHORTEST DISTANCE TRAVELLED BY AN
OBJECT FROM INITIAL POSITION TO
FINAL POSITION.
• IT HAS ONLY MAGNITUDE AND
DIRECTION. (VECTOR QUANTITY)
• CAN BE ZERO, POSITIVE OR NEGATIVE.

5. Question: An Object is moving in a circle of radius r. Calculate the distance and displacement?
(i) When it completes half the circle.
(ii) When it completes one full circle.
Answer: When object completes half the circle, it moves from A to B.
Distance= ½ x Circumference
= ½ x 2Πr
=Πr
Displacement=2 x Radius of Circle
= Diameter of Circle
= 2 r
A B
r
o

6. Question: A particle moves 3m north, then 4m east and finally 6m south. Calculate the distance travelled
and the displacement.
Answer: The particle starts from O and moves 3m North
(=OA), 4m East (=AB) and 6m South (=BC).
Distance Travelled= OA+AB+BC
=3m +4m+ 6m
=13m
Displacement = OC
= 𝑂𝐷2 + 𝐷𝐶2
= 𝐴𝐵2 + (𝐵𝐶 − 𝐵𝐷)2
= 62 + (6 − 3)2
=5m
3m
4m
A B
C
Do
N
S
E W

7. TYPES OF MOTION
BASED ON DISTANCE COVERED BASED ON DIRECTION
Uniform Motion
Non Uniform
Motion Linear MotionCircular Motion

8. UNIFORM MOTION
• WHEN AN OBJECT TRAVELS EQUAL
DISTANCES IN EQUAL INTERVALS OF TIME
• A CAR COVERS 4M IN FIRST 2SECONDS
AND ANOTHER 4M IN SECOND 2 SECONDS
AND SO ON
NON UNIFORM MOTION
• WHEN AN OBJECT TRAVELS UNEQUAL
DISTANCES IN EQUAL INTERVALS OF TIME.
• A CAR COVERS 8M IN FIRST 2 SECONDS
THEN 1M IN ANOTHER 2 SECONDS AND SO
ON

9. CIRCULAR MOTION
• WHEN AN OBJECT CONTINUOUSLY
CHANGING ITS DIRECTION IRRESPECTIVE
OF ITS UNIFORMITY.
LINEAR MOTION
• WHEN AN OBJECT MOVES IN A
PARTICULAR DIRECTION
IRRESPECTIVE OF ITS UNIFORMITY.

10. MEASURING THE RATE OF MOTION: SPEED
Which one is
faster??
A driver A takes
4 hours to cover
a distance of
200km from
Ambala to Delhi
A driver B takes 3
hours to cover a
distance of
200km from
Ambala to Delhi
Speed of a body is defined as the distance travelled by the
body in unit time.
Speed= Distance/ Time
If s is the distance travelled by a body
in time t, its speed is
v=s/t
SI unit of speed is m/s or m/s-1
Note: Speed is a scalar quantity. It has magnitude
only. The speed can be zero or positive.

11. TYPES OF SPEED:
Uniform
Speed
Non
uniform
Speed
When a body travels equal distances
in equal intervals of time, the speed
of the body is said to be uniform.
When a body covers unequal distances
in equal intervals of time, the speed of
the body is said to be non uniform.
Note: in most of cases, bodies move with non uniform speed. Therefore, we describe
their rate of motion in terms of average speed.

12. SPEED WITH DIRECTION: VELOCITY
Velocity of a body is the distance travelled by the body in unit time in a given direction.
or
Velocity of a body is the speed of the body in a particular direction.
Velocity = distance travelled in given direction/ time taken
or
Velocity = displacement/time → →
v = s / t
Where
→
v is velocity
→
s is displacement
Note: The unit of velocity is same as that of speed i.e. m/s
Velocity is a vector quantity whereas speed is scalar quantity

13. AVERAGE SPEED AND AVERAGE VELOCITY
During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the
speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car
reveals information about the instantaneous speed of your car. It shows your speed at a particular instant in
time (Instantaneous speed).
Average speed is a measure of the distance travelled in a given period of time.
Average velocity is a measure of displacement made by the object in a given period of time.
Suppose that during your trip to school, you travelled a distance of 5 miles and the trip lasted 0.2 hours (12 minutes). The
average speed of your car could be determined as

14. Question: Use the diagram to determine the average speed and the average velocity of the skier during
these three minutes.
Using v = s/t,
The skier has an average speed
of
(420 m) / (3 min) = 140 m/min
and
an average velocity of
(140 m, right) / (3 min) = 46.7
m/min

15. RATE OF CHANGE IN VELOCITY: ACCELERATION
• CHANGE IN SPEED OR DIRECTION OF AN OBJECT IN UNIT TIME
OR
• RATE OF CHANGE IN VELOCITY
Acceleration = change in velocity/ time
final velocity – Initial velocity
Acceleration = ---------------------------------------
Time
v – u
a = ---------
t
• Acceleration can be zero, negative or positive i.e. vector quantity
Note: change in velocity can be done by (a) change in speed (b) change in direction (c) both of them

16. TYPES OF ACCELERATION:
Uniform
Acceleration
Non-Uniform
acceleration
When a body travels with uniform or
constant velocity, the acceleration of
the body is said to be uniform.
When a body travels with non-uniform
velocity, the acceleration of the body is
said to be non uniform.
Note: In most of cases, bodies move with non uniform acceleration.

17. TYPES OF ACCELERATION:
Positive
Acceleration
Negative
acceleration
When acceleration is in the direction
of velocity, the acceleration of the
body is said to be positive.
Example: when car starts from rest
When acceleration is in the opposite
direction of velocity, the acceleration of
the body is said to be negative.
Example: when car stops to rest

18. MOTION IN GRAPHS
CONSTANT/UNIFORM
SPEED
CONSIDER A CAR MOVING WITH
A RIGHTWARD (+), CHANGING
VELOCITY - THAT IS, A CAR THAT IS
MOVING RIGHTWARD BUT SPEEDING UP
OR ACCELERATING.
NON UNIFORM SPEED
CONSIDER A CAR MOVING WITH
A CONSTANT, RIGHTWARD (+)
VELOCITY - SAY OF +10 M/S
Position vs Time Graphs

19. Let's begin by considering the position versus time graph below.
The diagram above shows this method being applied
to determine the slope of the line. Note that three
different calculations are performed for three
different sets of two points on the line. In each case,
the result is the same: the slope is 10 m/s.

20. Consider the graph below. Note that the slope is not positive but rather negative; that is, the line slopes
in the downward direction. Note also that the line on the graph does not pass through the origin.
Test your understanding of slope calculations by determining the slope of the line below.
Slope = -3.0 m/s

21. TEST YOUR UNDERSTANDING
Determine the velocity (i.e., slope) of the object as portrayed by the graph below.
The velocity (i.e., slope) is 4 m/s.

22. MOTION IN GRAPHS
CONSTANT VELOCITY
NOW CONSIDER A CAR MOVING
WITH A RIGHTWARD (+), CHANGING
COVERING EQUAL DISTANCES IN
EQUAL INTERVAL OF TIME.
(UNIFORM ACCELERATION)
CHANGING VELOCITY
NOW CONSIDER A CAR MOVING WITH A
RIGHTWARD (+), CHANGING VELOCITY
BUT COVERING UNEQUAL DISTANCES IN
EQUAL INTERVALS OF TIME.
( NON UNIFORM ACCELERATION)
CONSIDER A CAR MOVING WITH
A CONSTANT, RIGHTWARD (+)
VELOCITY - SAY OF +10 M/S.
Velocity vs Time Graphs

23. VARIOUS SCENARIOS OF MOTION

24. MORE IN V-T GRAPH

25. QUESTIONS ON GRAPHS!!!!!
Describe the motion depicted by the following velocity-time graphs. In your descriptions, make reference to
the direction of motion (+ or - direction), the velocity and acceleration and any changes in speed (speeding up
or slowing down) during the various time intervals (e.g., intervals A, B, and C).
Ans:
The object moves in the + direction at a constant speed -
zero acceleration (interval A).
The object then continues in the + direction while slowing
down with a negative acceleration (interval B).
Finally, the object moves at a constant speed in the +
direction, slower than before (interval C).

26. TEST YOUR KNOWLEDGE
Describe the motion depicted by the following velocity-time graphs. In your descriptions, make reference to
the direction of motion (+ or - direction), the velocity and acceleration and any changes in speed (speeding up
or slowing down) during the various time intervals (e.g., intervals A, B, and C).
The object moves in the + direction while
slowing down; this involves a negative
acceleration (interval A). It then remains at rest
(interval B). The object then moves in the -
direction while speeding up; this also involves a
negative acceleration (interval C).
The object moves in the + direction with a constant velocity and
zero acceleration (interval A). The object then slows down
while moving in the + direction (i.e., it has a negative
acceleration) until it finally reaches a 0 velocity (stops) (interval
B). Then the object moves in the - direction while speeding up;
this corresponds to a - acceleration (interval C).

27. DETERMINING THE SLOPE ON A V-T GRAPH
Let's begin by considering the velocity versus time graph below.
Slope = acceleration
For 1st second,
Slope = 4-0/1-0 = 4/1 = 4m/s2
For 2nd second
Slope = 8-4/2-1 = 4/1 = 4m/s2

28. TEST YOUR KNOWLEDGE
Consider the velocity-time graph below. Determine the acceleration (i.e., slope) of the object as
portrayed by the graph.
The acceleration (i.e., slope) is 4 m/s2.

29. DETERMINING THE AREA ON A V-T GRAPH
As learned in an earlier part of this lesson, a plot of velocity-time can be used to determine the acceleration of an
object (the slope). For velocity versus time graphs, the area bound by the line and the axes represents the
displacement.
Rectangle Triangle Trapezoid
Area = b • h Area = ½ • b • h Area = ½ • b • (h1 + h2)

30. PRACTICE TIME!!!!!
Area = ½ * b * h where b = 1 s
and h = 10 m/s
Area = ½ * (1 s) * (10 m/s)
Area = 5 m
Area = ½ * b * (h1 + h2)
Area = ½ * (2 s) * (10 m/s + 30 m/s)
Area = 40 m
Area = b * h
Area = (6 s) * (30 m/s)
Area = 180 m

31. EQUATIONS OF MOTION
THE KINEMATIC EQUATIONS ARE A SET OF FOUR
EQUATIONS THAT CAN BE UTILIZED TO PREDICT
UNKNOWN INFORMATION ABOUT AN OBJECT'S
MOTION IF OTHER INFORMATION IS KNOWN. THE
EQUATIONS CAN BE UTILIZED FOR ANY MOTION
THAT CAN BE DESCRIBED AS BEING EITHER A
CONSTANT VELOCITY MOTION (AN ACCELERATION
OF 0 M/S/S) OR A CONSTANT ACCELERATION
MOTION.
Eqaution (1) represent velocity-time relation
Eqaution (2) represents position-time relation
Equation (3) represents position-velocity relation

32. FIRST EQUATION OF MOTION

33. SECOND EQUATION OF MOTION

34. THIRD EQUATION OF MOTION

35. TEST YOUR KNOWLEDGE
1. An airplane accelerates down a runway at
3.20 m/s2 for 32.8 s until is finally lifts off the
ground. Determine the distance travelled
before take off.
Given: a = 3.20 m/s2
t = 32.8 s
u = 0 m/s
using second equation of motion,
s = ut + ½ at2
= 0 x 3.28 + ½ x 3.2 x 3.28 x
3.28
= 1721 m
2. 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?
Given: t = 2.60 s
u = 0 m/s
a = 9.8 m/s2
Using first and third equation of motion,
v = u + at
= 0 + 9.8 x 2.6
= 25.48 m
v2 – u2 = 2as
(25.48)2 – (0) 2 = 2 x 9.8 x s
649.23 – 0 = 19.6 x s
649.23/19.6 = s
34.1 m = s

36. FREE FALL AND THE ACCELERATION OF GRAVITY
A free falling object is an object that is falling under the sole influence of gravity. Any object that is being acted
upon only by the force of gravity is said to be in a state of free fall.

37. Characteristics of free falling objects:
• Free-falling objects do not encounter air resistance.
• All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated
as 10 m/s/s )
A position versus time graph for a
free-falling object is shown below.
A velocity versus time graph for a
free-falling object is shown below.
Equations of motion
with respect to free
fall

38. UNIFORM CIRCULAR MOTION
Did you ever stop to count the number of controls that cause acceleration in your car? At a minimum, you have
a gas pedal, a brake, and a steering wheel. That's right, the steering wheel produces an acceleration on your car,
because it causes you to change direction which changes your velocity, even if it doesn't change your speed.
That's what Uniform Circular Motion is all about - the changing of the velocity of an object as it travels in a
circle at a constant speed.

39. Uniform Circular Motion – an object moving in a circle at a constant speed.

40. CIRCULAR MOTION AND CENTRIPETAL
FORCE
An object moving in a circle is experiencing an acceleration. Even if moving around the perimeter of the circle
with a constant speed, there is still a change in velocity and subsequently an acceleration. This acceleration is
directed towards the centre of the circle.
In accord with Newton's second law of motion, an object which experiences an acceleration must also be
experiencing a net force. The direction of the net force is in the same direction as the acceleration. So for an
object moving in a circle, there must be an inward force acting upon it in order to cause its inward
acceleration. This is sometimes referred to as the centripetal force

41. TEST YOUR KNOWLEDGE
Q: A 900-kg car moving at 10 m/s takes a turn
around a circle with a radius of 25.0 m.
Determine the acceleration and the net force
acting upon the car.
Ans: To determine the acceleration of the car,
use the equation a = v2 / R. The solution is as follows:
a = v2 / R
a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
To determine the net force acting upon the car,
use the equation Fnet = m x a. The solution is as follows.
Fnet = m x a = (900 kg) x (4 m/s2)
Fnet = 3600 N
Q: Determine the centripetal force acting upon a 40-kg child
who makes 10 revolutions around the Cliff-hanger in 29.3
seconds. The radius of the barrel is 2.90 meters.
Ans:
Given: m = 40 kg; R = 2.90 m; T = 2.93 s (since 10 cycles takes
29.3 s).
First, find speed using
speed=(2𝜋R) / T = 6.22 m/s.
Then find the acceleration using
a = v2 / R = = (6.22 m/s)2 / (2.90 m) = 13.3 m/s2
Now use Fnet = m x a
Fnet = 40kg x 13.3m/s2 = 533 N.

42. “
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