A student is able to: - Plot and interpret displacement-time and velocity-time graphs - Determine an object's motion from the shape of the graphs, including whether it is at rest, moving with uniform or non-uniform velocity/acceleration - Calculate distance, displacement, velocity, and acceleration from displacement-time and velocity-time graphs - Solve problems involving linear motion using the equations of motion

1. 1

2. 2
•plot and interpret displacement- time and velocity-time graphs
•deduce from the shape of a displacement-time graph when
a body is:
i. at rest
ii. moving with uniform velocity
iii. moving with non-uniform velocity
•determine distance, displacement and velocity from a
displacement –time graph
•deduce from the shape of velocity- time graph when a body is:
i. at rest
ii. moving with uniform velocity
iii. moving with uniform acceleration
•determine distance, displacement velocity and acceleration
from a velocity–time graph
•solve problems on linear motion with uniform
.
A student is able to:
Learning OutcomesLearning Outcomes

3. 3
MOTION GRAPHS
DISPLACEMENT
– TIME GRAPH
VELOCITY –
TIME GRAPH
GRADIENT :
VELOCITY (v)
1: GRADIENT :
ACCELERATION (a)
2: AREA UNDER THE GRAPH :
DISPLACEMENT (s)

4. 4
DISPLACEMENT – TIME GRAPHDISPLACEMENT – TIME GRAPH
s / m
t /s
Gradient = s
t
s
t
= velocity
= m
s
= ms-1

5. 55
s / cm
t / s
10
2
v = gradient of s-t graph
= (10 – 0 ) cm
( 2 – 0 ) s
= 5 cm/s
Base on the graph, find the velocity of the object ‘s
Example 1Example 1

6. 6

7. 77
s : constant
v : zero
s
t
GRAPH 1GRAPH 1

8. 88
GRAPH 2GRAPH 2
s
t
s increases
uniformly
v uniform

9. 99
GRAPH 3GRAPH 3
Larger
increment in s
v increases
s
t

10. 1010
s
t
GRAPH 4GRAPH 4
Smaller
increment in s
v decreases

11. 11
GRAPH 5GRAPH 5
s
t
-v uniform
•Object moves at
opposite direction
s decreases
uniformly

12. 12
GRAPH 6GRAPH 6
s
t
Larger
decrement in s
-v ?

13. 1313
GRAPH 7GRAPH 7
s
t
-v ?
Smaller
decrement in s

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GRAPH VELOCITY - TIMEGRAPH VELOCITY - TIME
Gradient = v - u
t
v / ms-1
t /s
v - u
t
= ms-1
s
= ms-2
= acceleration

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Example 2Example 2
v / cms-1
t / s
10
2
a = gradient of v-t graph
= (10 – 0 ) cms-1
( 2 – 0 ) s
= 5 cm/s-2
Base on the graph, find the acceleration of the object ‘s

16. 16
Area under the graph = triangle area
= 1 ( v )(t)
2
= (ms-1
)(s)
v / ms-1
t /s
v
t
= displacement
= m
GRAPH VELOCITY - TIMEGRAPH VELOCITY - TIME

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v / ms-1
t /st
v
Area under the graph = rectangle area
= ( v )(t)
GRAPH VELOCITY - TIMEGRAPH VELOCITY - TIME

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ExampleExample
v (m/s)
t(s)
2
4
8
Base on the graph, calculate
the displacement of a
moving object.
Answer
Displacement = Area under
v-t graph
= ∆ area + area
=½ (8-2)m/s(4 s) + (2 m/s)(4s)
= 20 m

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20. 2020
v : constant
a : zero
v
t
GRAPH 1GRAPH 1

21. 2121
GRAPH 2GRAPH 2
v
t
v increases
uniformly
a uniform

22. 2222
GRAPH 3GRAPH 3
Larger
increment in v
a increases
v
t

23. 2323
v
t
GRAPH 4GRAPH 4
Smaller
increment in v
a decreases

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GRAPH 5GRAPH 5
v
t
-a uniform
v decreases
uniformly

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GRAPH 6GRAPH 6
v
t
Larger
decrement in v
-a ?

26. 2626
GRAPH 7GRAPH 7
v
t
-a ?
Smaller
decrement in v

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Equations of Linear MotionEquations of Linear Motion
Derived the equations
Problem solving

28. 28
Derived the equations
Acceleration = Gradient of
v-t graph
Rearrange [1]
Final velocity v = u+at …….[1a]
Displacement, s = area under v-t
graph
s = ½ ( u+v) t ……..[2]
Substitute [1a] into [2]
velocity
time
u
t
v
s = ½ ( u+ u+at) t
s = ut +½ at2
……...[3]
Substitute [1b] into [2]
Rearrange [1] again
s = ½ ( u+v) (v – u)
a
v2
= u2
+ 2as …….[4]
t = (v – u) ……..[1b]
a
a = (v – u) ……..[1]
t
v = (v + u) ……..[5]
2

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ConclusionConclusion
EquationEquation vv ss tt uu vv aa
11
22
33
44
55
66
a = (v – u)
t
v = s
t
s = ut +½ at2
v2
= u2
+ 2as
v = (v + u)
2
s = ½ ( u+v) t
√ √ √
√ √ √
√ √ √ √
√ √ √ √
√ √ √ √
√ √ √ √

30. 30
Where
s = displacement
u = Initial velocity
v = Final velocity
a = Constant acceleration
t = time interval
The Equations of Motion
atuv +=1.
u+v
2





S = t4.
asuv 222
+=3.
2
2
1
atuts +=2.

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Example 1Example 1
 A train is moving at a constant velocity ofA train is moving at a constant velocity of
10 m/s,. It is then uniformly accelerated in10 m/s,. It is then uniformly accelerated in
10 s to a velocity of 20 m/s. what is the10 s to a velocity of 20 m/s. what is the
distance traveled by the train?distance traveled by the train?
u = 10 m/s
v = 20 m/s
t = 10 s
s = (v + u) t
2
= (20 – 10)m/s(10s)
2
s = 50 m

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Example 2Example 2
 A cheetah accelerates from rest at 4 m/sA cheetah accelerates from rest at 4 m/s22
over a distance of 60m to catch a rabbit.over a distance of 60m to catch a rabbit.
What is the final speed?What is the final speed?
u = 0 m/s
a = 4 m/s2
s = 60 m
v2
= u2
+ 2as

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Example 3Example 3
 A new car start from rest and travel 400m in 16 s.A new car start from rest and travel 400m in 16 s.
a) What is its average acceleration duringa) What is its average acceleration during
this time?this time?
b) Calculate the final speed of the car.b) Calculate the final speed of the car.
c) How fast is this final speed in kmhc) How fast is this final speed in kmh-1-1

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