PHYSICS KSSM FORM 4

1. PREVIOUS LESSONS
►LIST BASE QUANTITIES & DERIVED
QUANTITIES. STATE S.I UNITS.
►DEFINITION SCALAR & VECTOR QUANTITIES
►TYPES OF GRAPH

2. FORCES AND MOTION
2.1 ANALYSING LINEAR MOTION

3. Learning Objective
►2.1.1 Describe the type of linear motion of
an object in the following states:
(i) stationary
(ii) uniform velocity
(iii) non-uniform velocity
►2.1.2 Determine:
(i) distance and displacement
(ii) speed and velocity
(iii) acceleration/ deceleration

4. INTRODUCTION
How fast ?
(Speed / velocity)
Does it change
its speed ?
(Acceleration / deceleration)
How would you
describe the
motion in word ?
How far does it
travel ?
(distance/displacement)
Information required:
How fast ?
(Speed / velocity)
How far does it
travel ?
(distance/displacement)

5. LINEAR MOTION NON-LINEAR MOTION

7. 1. Constant = uniform
2. Increasing velocity = acceleration
3. Decreasing velocity = deceleration
4. Zero velocity = at rest / object at
stationary
5. Negative velocity = object moves opposite
direction
6. Zero acceleration = constant velocity

8. • DISTANCE AND
DISPLACEMENT
• SPEED AND
VELOCITY
• ACCELERATION AND
DECELERATION
Learning area

9. DISTANCE AND DISPLACEMENT
How far is it
from his
house to
factory?
= total path length
= 200 m
= shortest path length
= 120 m

10. Comparison between distance
and displacement

11. SPEED AND VELOCITY
start
end
path
Distance =
Displacement =
Average Speed =
Average Velocity =
Time taken =

12. SPEED AND VELOCITY
start
end
path
Speed, v =
Velocity, v =
Distance
Time taken
Displacement
Time taken
scalar
vector

13. ACCELERATION AND DECELERATION
Velocity
increases
Constant
velocity
Velocity
decreases
Acceleration = Rate of change of velocity
= Change of velocity
Time
= final velocity – Initial velocity
Time
a = v – u
t
vector
m s-2
Velocity increases = acceleration
Velocity decreases = deceleration

14. Example
1. Every morning Amir walks to Ahmad’s house
which is situated 80 m to the east of Amir’s
house. They then walk towards their school
which is 60 m to the south of Ahmad’s house.
( a ) What is the distance traveled by Amir and his
displacement from his house?
( b ) If total time taken by Amir to travel from his
house to Ahmad’s house and then to school is
15 minutes, what is his speed and velocity?
Answer : distance travelled = 140 cm
dislacement = 100 cm
Answer : speed = 0.16 ms-1
velocity = 0.11 ms-1

15. ►2. A car moving round a big roundabout which has a
radius of 70m. Calculate :
► ( a ) the distance moved by the car
► ( b ) the displacement of the car
►Answer :
► ( a ) the distance moved by the car
distance moves = circumference
= 2 π r
= 2 x 22 x 70
7
= 440 m
► ( b ) the displacement of the car
displacement = 0 m.

16. ► 1. A cow moves 3 m to the east and then 4 m to the north. Find the :
( a ) total distance moved by the cow
( b ) displacement of the cow
2. A car start from point O and moves 50 m to the north in 60 seconds.
The car then moves 120 m to the west in 40 seconds. Finally it stops.
Calculate the :
( a ) total distance moved by the car
( b ) displacement of the car
( c ) velocity of the car
( d ) average speed of the car
( d ) speed of the car when it is moving to the north.
Exercise

17. 3. Saiful running in a race covers 60 m in 12s.
( a ) what is his speed in m/s ?
( b ) If he takes 40 s to complete the race, what is his
distance covered?
4. An aero plane fly's towards the north with a velocity 300
km/h in one hour. Then, the plane moves to the east with
the velocity 400 km/h in one hour.
( a ) what is the average speed of the plane?
( b ) what is the average velocity of the plane?
( c ) what is the difference between average speed and
average velocity of the plane?

18. Lesson 2
►

19. RELATING DISPLACEMENT,
VELOCITY , ACCELERATION
AND TIME
a. Using ticker tape
b. Using Equations
of Motion
Learning area

20. ticker timer
ticker tape
A.C. 50 Hz
50 dots made in 1 second
Carbon disc

21. DIRECTION OF MOTION OF TICKER TAPE

22. Time interval between two
adjacent dots = 1/50 s
= 0.02 s
1 tick = 0.02 s
dots
1 tick
Slow movement
faster movement
fastest movement

23. PREPARING A TAPE CHART (5 -TICKS STRIP)
0 5 10
First 5-tick
strip
2nd 5-tick strip
Velocity, v
(cm /s)
Time / s

24. INFERENCE FROM TICKER TAPE AND CHART
•Zero
acceleration
•constant
velocity
• Constant
acceleration
• Constant
deceleration

25. Aim : To use a ticker timer to identify the
types of motion
Discussion 2.3(A):
2. Spacing of the dots is further
means a higher speed.
Spacing of the dots is closer
means a slower speed.

26. Aim : To determine displacement, average
velocity and acceleration
Discussion 2.3(B):
1. Prepare a tape chart.
2. Determine average velocity using
v = Total displacement
time
3. Determine acceleration using
a = final velocity – initial velocity
time

27. TO DETERMINE THE AVERAGE VELOCITY
EXAMPLE
The time for each 5-tick strip = 5 x 0.02 s
= 0.1 s
Length / cm
Time / s
0
7
10
14
15
22
0.1
0.2
0.3
0.4
0.5
0.6
0.7
= (7 +10 +14 +15 +22 +14 +10) cm
= 92 cm
= 7 strips = 0.7 s
Total displacement
Total time taken
Average velocity = displacement
Time taken
= 92 / 0.7
= 131.4 cm s-1

28. TO DETERMINE THE ACCELERATION
EXAMPLE The time for each 10-tick strip = 10 x 0.02 s
= 0.2 s
5.8 / 0.2 =28 cm s-1
27.3 / 0.2 = 136.5
Initial velocity, u
Final velocity, v
acceleration = v-u
t
= (136.5 – 28) cm s-1
1.2 s
Length / cm
Time / s
0 0.2
0.4
0.6 1
1.2
Time takenTime taken
=(7-1 )strips
= 6 x 0.2 s
= 1.2 s
5.8
27.3
1.4
0.8
= 90.42 cm s-2

29. s = Displacement
u = Initial velocity
v = Final velocity
a = Constant
acceleration
t = Time interval
THE EQUATIONS OF MOTION
v u at 
21
2
s ut at 
2 2
2v u as 
2
u v
s t
 
 
 

30. EXAMPLE
A car travelling at a velocity 10 m s-1 due
north speeds up uniformly to a velocity of
25 m s-1 in 5 s. Calculate the acceleration
of the car during these five seconds
u = 10 m s-1 , v = 25 m s-1, t = 5 s, a = ?
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using v = u + at
25 = 10 + a(5)
5 a = 15
a = 3 m s-2
Don’t forget
the unit

31. EXAMPLE
A rocket is uniformly accelerated from
rest to a speed of 960 m s-1 in 1.5 minutes.
Calculate the distance travelled.
u = 0 m s-1 , v = 960 m s-1,
t = 1.5 x 60 = 90 s, s = ?
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using s = ½ (u + v)t
s = ½ (0 + 960) 90
= 43 200 m
What is the
unit ?

32. u =0 m s-1, a = 2.5 m s-2 , t = 10 s v = ? , s = ?
Using v = u + at
= 0 + (2.5)(10)
= 25 m s-1
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using s = ut + ½ at2
= 0(10) + ½ (2.5)(10)2
= 125 m

33. u = 25m s-1, v = 0 m s-1 , s = 50 m , a = ?
Using v 2 = u2 + 2as
0 = 252 + 2a (50)
0 = 625 + 100a
a = - 625
100
= - 6.25 m s-2
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
The negative sign shows
deceleration.

34. 1. A particle travelling due east at 2 m s-1
is uniformly accelerated at 5 m s-2
for 4 s. Calculate the displacement of
the particle.
u = 2 m s-1 , a = 5 m s-2, t = 4 s, s = ?
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using s = ut + ½ at2
s = 2(4) + ½ (5)(4)2
= 8 + 40
= 48 m
Exercise

35. 2. A trolley travelling with a velocity 2 m s-1
slides 10 m down a slope with a uniform
acceleration. The final velocity is 8 m s-1.
Calculate the acceleration.
u =2 m s-1 , v = 8 m s-1 , s = 10 m , a = ?
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using v2 = u2 + 2as
82 = 22 + 2 a (10)
20 a = 64 – 4
= 60
a = 3 m s-2

36. 3. Murali throws a ball vertically upwards
with a speed of 10 ms-1. What is the
acceleration of the ball one second after
leaving Murali’s hand?
u =10 m s-1 , v = 0 m s-1 , t = 1 s , a = ?
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using v = u + at
0 = 10 + a (1)
0 = 10 + a
a = `-10 ms-2
a = 10 ms-2 downwards

37. A car moves in a straight line from its stationary state
with a uniform acceleration. It achieve a velocity of a
120 ms-1 after moving through a distance of 100m.
Calculate:
( a ) the acceleration of the car
( b ) the time taken
( c ) the velocity when t = 3s
m
Answer :
( a ) 72 ms-2
( b ) 1.67 seconds
( c ) 216 m