TRAVEL GRAPHS
Higher Tier
DISTANCE TIME GRAPHS
Example
This distance-time graph shows the distance travelled by a walker in 60 seconds.
From O to A:
Walked 20 m in 20
seconds
From A to B:
Stationary for 10 seconds
From B to C
Walked 5 m in 30 seconds
Gradients
The gradient of the line OA
increase in y

increase in x
change in distance

change in time

On a distance-time graph
dis tan ce m

 velocity
time
s
Velocity from O to A = gradient OA =
20
=1 ms1
20
Velocity from A to B = gradient AB = 0 ms1
Velocity from B to C = gradient BC =
travelgraphs
5
1
 ms1.
30 6
©RSH 23-Mar-10
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TRAVEL GRAPHS
Higher Tier
VELOCITY TIME GRAPHS
This velocity-time graph shows the velocity of a vehicle at different times.
change in velocity v
  acceleration
change in time
t

Gradient at point A =

The area under the curve gives the distance travelled. In the diagram above, the shaded
region represents the distance travelled during the 10 seconds shown.

An estimate of this area can be found by dividing the region into trapeziums, working out
the area of each trapezium and then adding these together.
On a velocity-time graph
travelgraphs
©RSH 23-Mar-10
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TRAVEL GRAPHS
Higher Tier
Example
This table gives the velocity of a car at 5 second interval.
Time(s)
0
5
10
15
20
25
30
35
40
45
50
55
60
V (ms1)
0
8
12.5
15.5
18
21
24.5
30
31.5
32
30
25.5
20.5
a)
Draw a velocity-time graph
b)
At what instant was acceleration zero ? Explain.
c)
At what instant was acceleration greatest ? Explain.
d)
Find the acceleration when t = 35 s.
e)
What was the maximum speed ?
f)
Estimate the distance travelled between t = 40 and t = 60 seconds.
g)
Find the average speed between t= 40 and t = 6- seconds.
SOLUTION
a)
b) Acceleration = 0 when gradient of tangent to the curve is zero, at about t = 45s
c) Acceleration is greatest when gradient of tangent to the curve is steepest,
travelgraphs
©RSH 23-Mar-10
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TRAVEL GRAPHS
Higher Tier
at about t = 2s
d) When t = 35, gradient = 0.5 ms2.
e) Maximum speed = 32 ms1.
f)
Area = 567.5 m
Using the trapezium method,
5
5
5
5
A1 = 31.5  32  , A2 = 32  30  , A3 = 30  25.5 , A4 = 25.5  20.5
2
2
2
2
g) Average speed = distance travelled ÷ time taken = 567.5 ÷ 20 = 28.5 ms1.
travelgraphs
©RSH 23-Mar-10
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