Draw the distance-time graph
displacement / m
4
2
time / s
8
-2
10
18
RECAP
total distance / m
10
6
4
time / s
0
8
10
18
displacement / m
4
2
time / s
0
-2
8
10
18
POP-QUIZ
• These are distance-time graph and
displacement-time graph
• The slope of a distance-time graph represents
the speed of the object
• The slope of a displacement-time graph
represents the velocity of the object
• The different gradient / steepness of the slope
of a distance-time graph means that the object
is moving at a different speed
POP-QUIZ
• The steeper the slope of a distance-time graph,
the faster the speed of the moving object
• If the slope of a distance-time graph is of
constant steepness, it represents that the object
is moving at a constant/uniform speed
• If the slope of a distance-time graph is horizontal,
it represents that the object is stationary
• If the slope of a distance-time graph is not of
constant steepness, i.e. a curving slope, it
represents that the object is moving at a
non-uniform speed
INTERPRET THE GRAPH
total distance / m
object is stationary
object is travelling at non-uniform speed
D
C
B
A
object is travelling at uniform speed
0
increasing
speed
E
F
?
decreasing
speed
time / s
WHAT IS THE SPEED?
total distance / m
D
10
5
time / s
30
35
WHAT IS THE SPEED?
total distance / m
D
1 m/s ?
total distance travelled
---------------------------------total time taken
10
RISE = 5 m
5
RUN = 5 s
time / s
30
35
WHAT IS THE SPEED?
total distance / m
y2 – y1
-----------------------x2 – x1
1 m/s ?
INSTANTANEOUS SPEED = 3 m/s
D
10
9
X
6
5X
RISE = 3 m
RUN = 1 s
“SPEED” = 1 m/s ?
time / s
30
34
34.5 35
WHAT IS THE SPEED?
• Average speed :
“total distance travelled divided by total
time taken”
• Instantaneous speed :
“calculating the gradient of the tangent at
the point of interest”
POP-QUIZ
total distance / m
Point G
10
6
4
X
time / s
0
•
8
10
WHAT IS THE ASSUMPTION I MADE FOR SALLY’S JOURNEY?
SALLY’S SPEED IS UNIFORM BETWEEN 0-8 / 8-10 / 10-18 s
•
WHAT IS THE AVERAGE SPEED FOR THE JOURNEY?
0.56 m/s
•
18
WHAT IS THE INSTANTANEOUS SPEED AT POINT G?
1.00 m/s
SPEED–TIME GRAPH
• Distance-time graph / Displacement-time graph
• Speed-time graph / Velocity-time graph
• Acceleration-time graph
SPEED–TIME GRAPH
total distance / m
10
6
4
time / s
0
speed / m/s
8
speed / m/s
10
speed / m/s
18
speed / m/s
SPEED-TIME GRAPH
total distance / m
10
6
4
time / s
0
8
10
18
speed / m/s
1.0
0.5
time / s
0
8
10
18
SPEED-TIME GRAPH
total distance / m
14
10
6
4
0
speed / m/s
8
speed / m/s
10
18
speed / m/s
time / s
20 24
speed / m/s
SPEED-TIME GRAPH
total distance / m
14
10
6
4
time / s
0
8
10
18
20
24
speed / m/s
2.0
1.0
0.5
0
time / s
8
10
18
20
24
SPEED-TIME GRAPH
total distance / m
18
14
10
6
4
time / s
0
speed / m/s
8
10
speed / m/s
18
20
speed / m/s
24
28
speed / m/s
SPEED-TIME GRAPH
total distance / m
18
14
10
6
4
time / s
0
8
10
18
20
24
28
speed / m/s
2.0
1.0
0.5
time / s
0
8
10
18
20
24
28
AREA UNDER SPEED-TIME GRAPH
speed / m/s
2.0
1.0
0.5
time / s
0
8
10
18
20
24
28
total distance / m
18
14
10
6
4
time / s
0
8
10
18
20
24
28
POP-QUIZ
speed / m/s
1.0
0.5
0
time / s
8
18
26
total distance / m
18.0
14.0
time / s
4.0
0
8
18
26
graphical analysis of motion
From the velocity-time graph of a object, we can obtain 3
important pieces of information.
1.
Instantaneous velocity (read off y-axis)
2.
Acceleration = gradient of the graph
v–u
=
Δt
3.
Distance travelled
= area under the graph
= ½ (u + v) t
velocity/m s-1
v
Recall
Area of trapezium =
½ x (a + b) x height
u
t
time/
s
graphical analysis of motion
The motion of a car is described by the velocity-time graph
below. Find
i) The acceleration of the car between points A and B
velocity/m s-1
Solution
Acceleration = gradient of v-t graph
–5 – 20
= 8–4
20
A
= - 6.3 m/s2 (2sf)
4
-5
7
8
11 time/s
B
graphical analysis of motion
The motion of a car is described by the velocity-time graph
below. Find
ii) The average speed of the car
Solution
Average speed
velocity/m s-1
=
Total distance
Total time
=
Area under graph
Total time
20
X
Total Area = Area of X + Area of Y
= ½(7)(20) + ½(4+3)(5)
= 87.5 m
Average speed
= 87.5/11
= 8.0 m/s (2sf)
A
4
-5
7
8
Y 11 time/s
B
graphical analysis of motion
The motion of a car is described by the velocity-time graph
below. Find
iii) The total displacement of the car
velocity/m s-1
Solution
Displacement of car, s
= (+ve dist) – (–ve dist)
20
= Area of X – Area of Y
= ½(7)(20) - ½(4+3)(5)
= + 52.5 m
A
X
4
Or 52.5 m in the positive direction
-5
7
8
Y 11 time/s
B
Positive direction
Y
O
14 m
8m
X
A toy car moves from point X to point Y at a constant speed of 2 m/s,
passing the observer at O. Sketch the displacement-time graph for
the motion of the toy car.
displacement/m
To find time x (+ve part of graph)
Velocity = Gradient of the graph
0-8
-2 =
x
x =4
To find time y
Velocity = Gradient of the graph
- 14 - 8
-2 =
y
y = 11
8
x4
–14
y11 time/s
graphical analysis of motion
Displacement-time graphs
a) What motion is described by the displacement-time graph below?
displacement/m
•
Gradient is positive
throughout
•
Velocity is positive
•
Object moves in positive
direction
Larger gradient
Increasing velocity
Small gradient
Small velocity
time/s
The object is experiencing increasing velocity in the positive direction
graphical analysis of motion
Displacement-time graphs
b) What motion is described by the displacement-time graph below?
displacement/
m
Smaller gradient
Decreasing velocity
Large gradient
•
Gradient is positive
throughout
•
Velocity is positive
•
Object moves in positive
direction
High velocity
time/
s
The object is experiencing decreasing velocity in the positive direction
graphical analysis of motion
Displacement-time graphs
c) What motion is described by the displacement-time graph below?
displacement/
m
Small gradient
Small velocity
•
Gradient is negative
throughout
•
Velocity is negative
•
Object moves in
negative direction
steeper gradient
Increasing velocity
time/
s
The object is experiencing increasing velocity in the negative direction
graphical analysis of motion
Displacement-time graphs
d) What motion is described by the displacement-time graph below?
displacement/
m
Steep gradient
•
Gradient is negative
throughout
•
Object moves in negative
direction
•
Negative gradient does
not imply object moving
towards observer
•
Displacement approach
zero either from +ve or –
ve direction implies
object moving towards
observer
Large velocity
gentler gradient
Decreasing velocity
time/
s
The object is experiencing decreasing velocity in the negative
direction
Deceleration vs Acceleration
•Let's suppose we define East as positive. Consider each of the following
situations with this sign convention for all.
Suppose we have a car traveling West and speeding up.
1. Is velocity negative or positive? Answer: negative
2. Is acceleration negative or positive? Answer: negative
Suppose we have a car traveling West and slowing down.
1. Is velocity negative or positive? Answer: negative
2. Is acceleration negative or positive? Answer: positive
Suppose we have a car traveling East and speeding up.
1. Is velocity negative or positive? Answer: positive
2. Is acceleration negative or positive? Answer: positive
Suppose we have a car traveling East and slowing down.
1. Is velocity negative or positive? Answer: positive
2. Is acceleration negative or positive? Answer: negative