Motion along-a-straight-line

1. Motion along a straight line
Standard 9

2. Person outside the bus
The bus moved away from the tree
The person is comparing the position of the bus with respect to the position of the tree
Reference (or origin) is position of the tree
0 5 10

3. Person inside the bus
The tree moved away from the bus.
The person is comparing the position of the tree with respect to the position of the bus.
Reference (or origin) is position of the bus.
0510

4. Motion is relative
Both the observations are correct. The difference is what is taken as the origin.
Motion is always relative. When one says that a object is moving, he/she is comparing
the position of that object with another object.
Motion is therefore change in position of an object with respect to another object over
time.
Kinematics studies motion without delving into what caused the motion.

5. Direct Path (1.1 km)Actual Path (2 km)
Q. How much distance do you have to travel to reach school?
Q. If you were to draw a straight line between your house and school, what would be
the length of that line?

6. Direct Path (1.1 km)Actual Path (2 km)
Q. How much distance do you travel in one round trip to the school?
Q. After one trip how far away are you from your home?

7. Distance and Displacement
Distance = length of the actual path taken to go from
source to destination
Displacement = length of the straight line joining the
source to the destination or in other words the length
of the shortest path

8. Checkpoint
Suppose it was given that the person started by point A and walked in a straight line for
5 km. Can you calculate the end point of his/her journey?
A
No, the person could be anywhere on the
circle of 5 km radius.
Unless we know the direction of the motion
we cannot calculate the end point of the
journey.

9. Sample Problem
Rohit and Seema both start from their house. Rohit walks 2 km to the east while Seema
walk 1 km to the west and then turns back and walks 1 km.
Distance travelled by them is the same (2 km)
Is their displacement also the same?
No – Seema is back home and her displacement is 0 m.
This is because direction of motion is different in both cases.
You require both distance and direction to determine displacement.

10. Sample Problem
A B
C
Distance AB = 3 km due East
Distance BC = 4 km due North
What is the distance travelled by a person who moves from
A to C via B?
What is the displacement? What is the direction of the
displacement?
Distance travelled = 7 km, Displacement = 5 km from A towards C.

11. Rate of Motion
Distance travelled per unit time or the displacement
per unit time.
𝑠𝑝𝑒𝑒𝑑 =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒
𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐𝑜𝑛𝑑
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒
𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐𝑜𝑛𝑑
When an object is travelling along a straight line its
velocity is equal to its speed.

12. Sample Problem
r = 100 m
Distance = 2 ∗ Π ∗ 𝑟 ∗ 𝑙𝑎𝑝𝑠 = 6280 m, Displacement = 0 m
Speed =
𝑑𝑖𝑠𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒
𝑚/𝑠 =
6280
125.6
𝑚/𝑠 = 180 km/hr
The adjoining figure shows a Formula 1 racing
track. A driver is did 10 laps, what is the
distance travelled by the driver at the end of
the race?
What is the displacement?
If the driver took 125.6 seconds to complete
the laps, what is his speed and velocity in
km/hr?

13. Uniform Motion
A distance – time graph represents the
distance travelled with respect to time.
When an object covers equal distance in
every time interval, it is said to be having
uniform motion.
In an uniform motion, the speed of the
object remains constant.
A stationary body is also an example of
uniform motion
25
20
15
10
5
0
10 20 30 40 500
Distance(m)
Time (s)
Distance – Time graph

14. 1.25
1.0
0.75
0.5
0.25
0
10 20 30 40 500
Speed(m)/s
Time (s)
Speed – Time graph
Area of shaded region = 0.5 * 40 = 20mDistance travelled = 20 m
Velocity – Time graph
25
20
15
10
5
0
10 20 30 40 500
Distance(m)
Time (s)
Distance – Time graph

15. 1.25
1.0
0.75
0.5
0.25
0
10 20 30 40 500
Velocity(m)/s
Time (s)
Velocity – Time graph
Uniform Motion
Acceleration = 0 m/s2
1.25
1.0
0.75
0.5
0.25
0
10 20 30 40 500
Velocity(m)/s
Time (s)
Velocity – Time graph
Non-uniform Motion
Acceleration = 0.125 m/s2
Uniform and Non-Uniform Motion

16. Rate of Change of Velocity
Rate of change of velocity
acceleration = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡𝑖𝑚𝑒
meter/second2
A body is said to be accelerating if there is a change
in velocity.
Velocity has magnitude and direction. A body has
acceleration when either of them changes.

17. 1.25
1.0
0.75
0.5
0.25
0
10 20 30 40 500
Velocity(m)/s
Time (s)
Velocity – Time graph
Uniform Acceleration
Acceleration = 0.125 m/s2
1.25
1.0
0.75
0.5
0.25
0
10 20 30 40 500
Velocity(m)/s
Time (s)
Velocity – Time graph
Non-uniform Acceleration
Uniform and Non-Uniform
Acceleration

18. Sample Problem
25
20
15
10
5
0
10 20 30 40 500
Velocity(m/s)
Time (s)
A
B
C
D
Which object has the maximum
acceleration?
Which object has no acceleration?
How much distance is covered by
object D in 20 seconds?
Explain the motion represented by D.
Given an example of such a motion
in real life.

19. 1st Equation of Motion
v
u
0
t0
Velocity(m)/s
Time (s)
Velocity – Time graph
Uniform Acceleration
Initial velocity = u
Final velocity = v
Time = t
Acceleration = a
Displacement = s
Acceleration = Rate of change of velocity
𝑎 =
𝑣 − 𝑢
𝑡
𝑚/𝑠2
𝑎𝑡 = 𝑣 − 𝑢
𝒗 = 𝒖 + 𝒂𝒕 𝒎/𝒔

20. 2nd Equation of Motion
v
u
0
t0
Velocity(m)/s
Time (s)
Velocity – Time graph
Uniform Acceleration
Initial velocity = u
Final velocity = v
Time = t
Acceleration = a
Displacement = s
Displacement = Area under the line
𝑠 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
𝑠 = 𝑢𝑡 +
1
2
𝑡 ∗ 𝑣 − 𝑢 𝑚
But
(𝑣−𝑢)
𝑡
= 𝑎 or 𝑣 − 𝑢 = 𝑎𝑡
𝒔 = 𝒖𝒕 +
𝟏
𝟐
𝒂𝒕 𝟐
𝒎

21. 3rd Equation of Motion
v
u
0
t0
Velocity(m)/s
Time (s)
Velocity – Time graph
Uniform Acceleration
Initial velocity = u
Final velocity = v
Time = t
Acceleration = a
Displacement = s
Displacement = Area under the line
𝑠 = 𝑎𝑟𝑒𝑎 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚
𝑠 =
1
2
𝑢 + 𝑣 ∗ 𝑡 𝑚
But
(𝑣−𝑢)
𝑎
=t
𝑠 =
1
2
𝑢 + 𝑣 ∗
(𝑣 − 𝑢)
𝑎
𝑚
∴ 2𝑎𝑠 = 𝑢 + 𝑣 ∗ 𝑢 − 𝑣
∴ 2𝑎𝑠 = 𝑢2 − 𝑣2
𝒗 𝟐 = 𝒖 𝟐 + 𝟐𝒂𝒔