1. Motion along a straight line
Standard 9

2. Person outside the bus 5 10 The bus moved away from the tree5 10
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

3. Person inside the bus 10 5 The tree moved away from 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.

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. Q. How much distance do you have to travel to reach school?
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. Q. How much distance do you travel in one round trip to the school?
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?
No, the person could be anywhere on the circle of 5 km radius. A
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. Distance travelled = 7 km, Displacement = 5 km from A towards C.
Sample Problem 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? A B
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
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 seconds to complete the laps, what is his speed and velocity in km/hr?
Distance = 2∗Π∗𝑟 ∗𝑙𝑎𝑝𝑠 = 6280 m, Displacement = 0 m
Speed = 𝑑𝑖𝑠𝑎𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 𝑚/𝑠 = 𝑚/𝑠 = 180 km/hr

13. Uniform Motion 25 20 15 10 5 30 40 50
Distance (m)
Time (s)
Distance – Time graph
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

14. Velocity – Time graph Distance travelled = 20 m25 20 15 10 5 30 40 50
Distance (m)
Time (s)
Distance – Time graph
1.25
1.0
0.75
0.5
0.25 10 20 30 40 50
Speed (m)/s
Time (s)
Speed – Time graph
Distance travelled = 20 m
Area of shaded region = 0.5 * 40 = 20m

15. Uniform and Non-Uniform Motion
1.25
1.0
0.75
0.5
0.25 10 20 30 40 50
Velocity (m)/s
Time (s)
Velocity – Time graph
Uniform Motion
Acceleration = 0 m/s2
1.25
1.0
0.75
0.5
0.25 10 20 30 40 50
Velocity (m)/s
Time (s)
Velocity – Time graph
Non-uniform Motion
Acceleration = m/s2

16. 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. Uniform and Non-Uniform Acceleration
1.25
1.0
0.75
0.5
0.25 10 20 30 40 50
Velocity (m)/s
Time (s)
Velocity – Time graph
Uniform Acceleration
Acceleration = m/s2
1.25
1.0
0.75
0.5
0.25 10 20 30 40 50
Velocity (m)/s
Time (s)
Velocity – Time graph
Non-uniform Acceleration

18. Sample Problem Which object has the maximum acceleration?25 20 15 10 5 30 40 50
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 Initial velocity = u Final velocity = vt
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/ 𝑠 2
𝑎𝑡=𝑣 −𝑢
𝒗=𝒖+𝒂𝒕 𝒎/𝒔

20. 2nd Equation of Motion Initial velocity = u Final velocity = vt
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 𝑡 ∗ 𝑣 −𝑢 𝑚
But (𝑣−𝑢) 𝑡 =𝑎 or 𝑣 −𝑢 =𝑎𝑡
𝒔=𝒖𝒕+ 𝟏 𝟐 𝒂 𝒕 𝟐 𝒎

21. 3rd Equation of Motion Initial velocity = u Final velocity = vt
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= 𝑢+𝑣 ∗𝑡 𝑚
But (𝑣−𝑢) 𝑎 =t
𝑠= 𝑢+𝑣 ∗ (𝑣−𝑢) 𝑎 𝑚
∴2𝑎𝑠= 𝑢+𝑣 ∗ 𝑢 −𝑣
∴2𝑎𝑠= 𝑢 2 − 𝑣 2
𝒗 𝟐 = 𝒖 𝟐 +𝟐𝒂𝒔