This document discusses speed and velocity. It defines speed as the distance moved per second, which is a scalar quantity measured in m/s. Velocity is defined as the rate of change of displacement, which is a vector quantity also measured in m/s. The document provides examples of calculating speed and velocity from distance and time measurements. It also includes graphs showing displacement and velocity over time to illustrate constant, accelerating, and decelerating motion.

1. Speed and velocity

2. Learning objectives
Students will be able to
• Define speed
• Define velocity
• Calculate speed and velocity

3. WARM UP

4. Speed
• Distance moved per second or rate of
change of distance
• It is scalar quantity
• Unit-m/s
• Average speed=Distance travelled/time
taken

5. Velocity
• Velocity measures rate of change of
displacement
• Average velocity=Displacement/time taken
• It is a vector quantity.
• Unit- m/s

6. CONTINOUS ASSESSMENT
1.What does the speedometer of a car measure?
2.A car travels a distance of 2oom and returns to the
original position after 5min.What is the speed and
velocity of the car?
3.What is the eqn. for speed?
4.What is the eqn. for time taken?

8. Final assessment
1.a)Define speed
b) Define velocity
(page42)

9. HOME WORK
• Page.33
Example.2

10. GRAPHS
Displacement –time graph
a) Body at rest
t
S
t(s) 0 1 2 3 4
S(m) 2 2 2 2 2

11. b)Body moving with steady velocity
t
s
t(s) 0 1 2 3 4
S(m) 0 5 10 15 20
Displacement
Gradient=
Time
Gradient of displacement –time graph gives
velocity. Steeper the line ,larger the velocity.

12. c)Accelerating body
t 0 1s 2s 3s 4s
s 0m 1m 4m 9m 16
a=2m/s²
t
s
Gradient at different time instants are different, as the
velocity is changing at every instant.
Gradient at any instant gives the instantaneous
velocity

13. d)Decelerating body
t
s

14. Velocity –time graph
a) Body moving with steady velocity
t
v

15. b)Body moving with steady acceleration
t
v
t(s) 0 1 2 3 4
v(m/s) 0 5 10 15 20
acceleration
Changeinvelocity
• Gradient=
Time
• Gradient of velocity –time graph gives
acceleration.
• Steeper the line ,larger the acceleration.


16. c)Body moving with steady deceleration
t
v
t(s) 0 1 2 3 4
v(m/s) 20 15 10 5 0
Changeinvelocity
• Gradient=
• Gradient of velocity –time graph gives
acceleration.
• Steeper the line ,larger the deceleration.
acceleration deceleration
Time
 /

17. Area under the velocity –Time graph gives the
displacement
Calculate the total displacement of
the given body using the given
velocity time graph.

18. OA-accelerated motion; Displacement = 100m
AB-Steady velocity ; Displacement = 400m
BC—decelerated motion; Displacement = 50m
Total displacement = 100+400+50=550m

19. Page43. Qn . 8

20. a) acceleration= gradient of OA
=change on y/change on x
=y2-y1/X2-X1 =2-0/3-0
=2/3 =1.5m/s²
b) Decelerates, its velocity becomes zero and it
reaches the maximum height.
c)At point C the velocity of the lift is zero. The
height=Displacement=area under OABC
= ½ × 3 × 2 + 6 × 2 + ½ × 3 × 2
=3+12+3=18m.
d)The lift starts moving down. First it accelerates in
the downward direction.Then it starts decelerating
and come to rest e) 18-6=12m

21. PAGE 106