The document discusses analyzing motion graphs through displacement-time graphs and velocity-time graphs. It provides examples of how to interpret these graphs: - Displacement-time graphs can be used to determine velocity from the gradient and qualitative motion details. Velocity-time graphs allow determining acceleration from the gradient and displacement from the area under the graph. - Examples show calculating displacement, velocity, acceleration and average values from given motion graphs through analyzing gradients and areas. - Motion graphs can qualitatively and quantitatively describe an object's motion by showing properties like constant/changing velocity and acceleration, as well as calculating kinematic values over different time periods.

1. 2.2
ANALYSING MOTION
GRAPHS

2. Analying Motion Graphs
Data of the motion can be presented in
the form of a graph called a motion graph

3. THE DISPLACEMENT-TIME GRAPH
Gradient = change of displacement
time
= velocity
S / m
t / sO
A
B
Positive velocity
means moving
forward
negative velocity
means moving
backward

4. THE DISPLACEMENT-TIME GRAPH
s / m
t / s
The gradient is
positive and constant
Moving forward with a
constant velocity
O
A
B
OA
AB The gradient is
negative and constant
Moving backward with
a constant velocity

5. s / m
t / s
The gradient is zero
zero velocity ie. at rest
THE DISPLACEMENT-TIME GRAPH
Moving forward with a
constant velocity and reach
the reference point after 3 s
s / m
t / s
3

6. s / m
t / s
The gradient is positive
and increasing
Moving forward with
an acceleration
s / m
t / s
The gradient is positive
and decreasing
Moving forward with a
deceleration

8. THE VELOCITY-TIME GRAPH
Gradient = change of velocity
time
= acceleration
t / s
V/ m s-1
O
A
B

9. THE VELOCITY-TIME GRAPH
The gradient is
negative and constant
Uniform deceleration
AB
BC The gradient is
zero
Zero acceleration i.e.
uniform velocity
v / m s-1
t / sO
A
B C

10. summary
Displacement-
time graph
Velocity-time
graph
gradientRepresents
velocity
Represents
acceleration
Positive
gradient
Moves forward
Negative
gradient
Moves backward
Zero
gradient
stationary
Straight lineUniform velocity Uniform acceleration
acceleration
deceleration
Uniform velocity
Area under graphNo significance Represents distance
Intercept time-axis
Passing through
the reference point
Object stops

11. evaluation
1. Which of the following graphs shows a body moving with
decreasing acceleration ?
Displacement
Time
Displacement
Time
Velocity
Time
Velocity
Time
A
B
C
D

12. evaluation
1. Which of the following graphs shows a body moving with
decreasing acceleration ?
Displacement
Time
Displacement
Time
Velocity
Time
Velocity
Time
A
B
C
D

13. 2. Figure below shows the displacement- time graph of a boy.
Displacement /m
Time
A
B
C
D
0
The motion of the boy in section AB and BC is
A
B
C
Section AB Section BC
Moving backward at rest
Moving backward uniform velocity
Moving forward at rest
deceleration uniform velocity

14. 2. Figure below shows the displacement- time graph of a boy.
Displacement /m
Time
A
B
C
D
0
The motion of the boy in section AB and BC is
A
B
C
Section AB Section BC
Moving backward at rest
Moving backward uniform velocity
Moving forward at rest
deceleration uniform velocity

15. 3. What information is represented by the gradient and area
under a velocity- time graph ?
A
B
C
D
Gradient Area under graph
Acceleration Average velocity
Acceleration displacement
Average velocity displacement
Average velocity acceleration

16. 3. What information is represented by the gradient and area
under a velocity- time graph ?
A
B
C
D
Gradient Area under graph
Acceleration Average velocity
Acceleration displacement
Average velocity displacement
Average velocity acceleration

17. 4. Figure below shows the velocity- time graph of a lorry.
A
B
C
Which of the followings is the correct description of its motion ?
Moving forward with constant velocity , at rest
and then moves backward
Accelerates, at rest and then decelerates.
Accelerates, moving with constant velocity
and then decelerates.
velocity
Time0
A B
C

18. 4. Figure below shows the velocity- time graph of a lorry.
A
B
C
Which of the followings is the correct description of its motion ?
Moving forward with constant velocity , at rest
and then moves backward
Accelerates, at rest and then decelerates.
Accelerates, moving with constant velocity
and then decelerates.
velocity
Time0
A B
C

19. 5. Figure below shows the velocity- time graph of a car.
A
B
C
The acceleration-time
graph for the car is
Velocity/m s-1
Time/s0
2 3 6
acceleration
time
acceleration
time
acceleration
time
acceleration
time
D

20. 5. Figure below shows the velocity- time graph of a car.
A
B
C
The acceleration-time
graph for the car is
Velocity/m s-1
Time/s0
2 3 6
acceleration
time
acceleration
time
acceleration
time
acceleration
time
D

21. To determine displacement,
velocity and acceleration from
motion graph

22. Example The graph shows the motion of a moving particle.
(a) What is the displacement of the
particle from the starting point
just before it moves with a uniform
velocity.
(b) Calculate the velocity
of the particle in the
first 20 s.
O
0 10 20 30 40 50 60
Time / s
20
40
60
80
100
Displacement /m
A
B
(c) Calculate the average
velocity
20 m
Velocity = gradient AB
Average Velocity
= displacement
time
= 80 + (-60) = 0.333 m s-1
60
= (100 – 20)m
20 s
= 4 m s-1
S1 = 80 m, S2 = -60m,
S3= 0 m

23. Example The graph shows the motion of a moving particle.
(a) What is the velocity of the particle
from t=30 s to t= 60 s ?
(b) Calculate the average
velocity
(c) Calculate the average
speed
Velocity = gradient
(c) Average speed
= total distance
time
= 25+ 15 = 0.667 m s-1
60
0
10 20 30 40 50 60
Time / s
-5
5
10
15
Displacement /m
B
-10
(b) Average Velocity
= displacement
time
= (-25 )+15 = -0.167 m s-1
60
(30,-10)
(60, 5)= 5-(-10) m
(60-30) s
= 0.5 m s-1
S1 = -10-15 = -25 m
S2 = 5 –(-10) = 15m

24. Example The graph shows the motion of a motorcycle.
(a) Calculate the deceleration of the
motorcycle.
(b) Calculate the average
velocity of the
motorcycle.
velocity/m s-1
Deceleration = gradient CD
= 0 - 20 m s-1
60 – 40 s
= -1 m s-2
Average Velocity
= Displacement
time
(40,
20)
(60, 0)
0 10 20 30 40 50 60
Time / s
5
10
15
20
A B
D
S1
C
S3
S2
S1= (10)(10) = 100 m
S2= 1 (10+ 20)30 = 450m
2
S3= 1 (20)(20) = 200m
2
= 100 + 450 + 200
60
= 12.5 ms-1

25. S2
S1
Example The graph shows the motion of a motorcycle.
(a) Calculate the average
velocity of the motorcycle
(b) Calculate the average
speed of the motorcycle.
= 150 + 150 + 25 m
60 s
= 5.417 ms-1
(a) Average Velocity
= Displacement
time
S1= ½ (20)(15) =150 m
velocity/m s-1
0
10 20 30 40 50 60
Time / s
-5
5
10
15
-10
S3
S2= ½ (30)(-10) =-150 m
S3= ½ (10)(5) 25 m
= 150 + (-150)+ 25 m
60 s
= 0.417 m s-1
(b) Average speed
= Total distance
time

26. Summary
Displacement-time graph
 determine velocity from the gradient
 area under graph = no significance
Velocity -time graph
 determine acceleration from the gradient
 determine displacement from the area
under graph
Average velocity = displacement / time
Average speed = distance / time

27. Exercise
1. The displacement-time graph
shows the motion of an object.
( a ) briefly describe the motion of
the object represented by
AB,BC, CD and DE.
( b ) ( i )Find the displacement
after 20 s
( ii ) time taken to move
from 35m from the
origin
( c ) Calculate the average velocity
in each of these time intervals.
( I ) 0 s - 5 s
( ii ) 5 s – 10 s
( iii ) 10 s 20 s
( iv ) 28 s – 20 s
The displacement-time graph
shows the motion of an object.

28. The displacement-time graph shows
the motion of an object.
( a ) briefly describe the motion of the
object represented by
AB = object is at rest 20 m from starting
point
BC = object moves 15 m forward with
positive velocity
CD = object moves forward 10 m with
positive velocity
DE = object moves backward 45 m with
negative velocity and return to its
starting point
( b ) ( i )Find the displacement after
20 s
25 m
( ii ) time taken to move from 35m
from the origin
10 s
( c ) Calculate the average
velocity in each of
these time intervals.
( I ) 0 s - 5 s
Answer = 0 ms-1
( ii ) 5 s – 10 s
Answer = 3 ms-1
( iii ) 10 s - 20 s
Answer = 1 ms-1
( iv ) 28 s – 20 s
Answer = - 5.63 ms-1

29.  2. The velocity – time graph
shows the motion of an object.
( a ) Find the velocity when
( i ) t = 5 s
( ii ) t = 18 s
( b ) Find the displacement at the
given time intervals
( i ) 0s – 5 s
( ii ) 5 s – 10 s
( iii ) 18 s – 24 s
( c ) Find the acceleration of the
object at
( I ) t = 1 s
( ii ) t = 7 s
( iii ) t = 12 s
( iv ) t = 16 s
( v ) t = 20 s

30.  2. The velocity – time graph
shows the motion of an object.
( a ) Find the velocity when
( i ) t = 5 s
( ii ) t = 18 s
( b ) Find the displacement at the
given time intervals
( i ) 0s – 5 s
( ii ) 5 s – 10 s
( iii ) 18 s – 24 s
( c ) Find the acceleration of the
object at
( i ) t = 1 s
( ii ) t = 7 s
( iii ) t = 12 s
( iv ) t = 16 s
( v ) t = 20 s
answer :
( a ) ( i ) velocity, v = 10 ms-1
( ii ) velocity = 18 ms-1
( b ) ( i ) 25 m ( ii ) 50 m ( iii ) 54 m
( c ) ( i ) a = 2 ms-2
( ii ) a = 0 ms-2
( iii ) a = -1.25 ms-2
( iv ) a = 3.25 ms-2
( v ) a = -3 ms-2

31. LESSON 2
Interpreting and
Sketching Graphs

32. Figure 2.24 shows the displacement-time graph of an
object in linear motion.

33. (a) Determine the velocity of the object at each stage of its
motion.
(b) Interpret the displacement-time graph in Figure 2.24 to
sketch a velocity-time graph.
ANSWER:
(a) Velocity = Gradient of displacement-time
graph

35. Figure 2.26 shows the velocity-time graph that is
plotted based on the linear motion of a car driven by
Encik Kassim. He drives his car at a speed of 30 m/s
and applies the brakes when he sees an obstruction
on the road.
Interpret the velocity-time graph of the motion of his
car and sketch;

36. (a) displacement-time graph
Displacement = AREA UNDER GRAPH
Total displacement after 10 s,
s = 180 + 60 = 240 m

37. (b) acceleration-time graph
Acceleration = GRADIENT OF THE GRAPH

38. Exercise
Formative Practice 2.2
Page : 45 TextBook