Topic 1: Travel graphs
Distance-Time graphs
Distance-Time graphs are used to illustrate journeys.
Time is always on the horizontal axis.
Distance is always on the vertical axis.
Remember:
speed 
dis tan ce
time
The speed of the object is found by calculating the gradient of the graph.
The faster the object is travelling the steeper the graph will be and therefore the
greater the gradient will be.
When the distance-time graph is linear (a straight line) the speed of the object is
constant,
A horizontal line has zero gradient, therefore the object has zero speed, i.e. it is not
moving.
Examples
See examples page 147 mathematics for AQA higher tier
Speed-Time graphs
Speed-Time graphs are also used to illustrate journeys.
Time is always on the horizontal axis.
Speed is always on the vertical axis.
acceleration 
speed
time
The acceleration of the object is found by calculating the gradient of the graph.
The greater the acceleration of the object the steeper the graph will be and therefore
the greater the gradient will be.
When the speed-time graph is linear (a straight line) the acceleration is constant.
A horizontal line has zero gradient so therefore the object has zero acceleration i.e.
the object is moving at a constant speed.
If the gradient is positive the object is accelerating.
If the gradient is negative the object is decelerating.
The area enclosed by a speed-time graph represents the distance travelled by the
object.
Example
The graph shows Jane’s speed plotted against time during a run.
Speed (m/s)
5
4
3
2
1
0
0
20
50
100
Time (seconds)
a)
b)
c)
d)
Give a brief description of Jane’s run.
Calculate Jane’s acceleration during the first 20 seconds of her run.
Calculate Jane’s deceleration during the last 50 seconds of her run.
How far did Jane run?
Solution
a) Jane accelerates at a constant rate for the first 20 seconds of her run. She
then runs at a constant speed of 4 m/s for 30 seconds before slowing at a
constant rate and stopping after 100 seconds.
b) Acceleration = speed ÷ time
= 4 ÷ 20
= 0.2 m/s²
c) Deceleration = speed ÷ time
= 4 ÷ 50
= 0.08 m/s²
d) The area under the graph represents the distance travelled.
Method 1: Finding the area of trapezium.
Distance travelled= ½(30 + 100) x 4
= 260m
Method 2: split graph into 2 triangles and a rectangle.
Distance travelled while accelerating= ½(20 x 4)
= 40m
Distance travelled at constant speed= 30 x 4
= 120m
Distance travelled while decelerating= ½(50 x 4)
= 100m
Total distance= 40 + 120 + 100 = 260m
Displacement-Time graphs
Displacement is distance in a specified direction.
This means displacement is a vector which gives the direction as well as the distance
travelled.
e.g 10km is a distance. 10 km north is a displacement.
In a displacement- time graph time is always along the horizontal axis and displacement
is always along the vertical axis.
Example
Positive direction
30m
20m
10m
Jane is competing in a bean bag race at a primary school. Starting at the basket she
collects the bean bags one at a time and puts them in the basket. Then she runs to the
finishing line. Jane runs at a constant speed of 2m/s.
Draw a displacement-time graph for Jane’s race.
Displacement from basket
(metres)
Solution
40
30
20
10
0
-10 0
-20
10
20
35
-30
-40
Time (seconds)
50
65
A positive displacement means that Jane is on the bean bag side of the basket.
A negative displacement means that Jane is on the finishing line side of the basket.
There are two abbreviations for displacement-time graphs you may come across. They
are x-t graph and a (t,x) graph. (x is the abbreviation often used for displacement).
Velocity
Velocity is the rate of change of displacement with time
The velocity can also be said to be speed in a certain direction.
The velocity of an object is found by calculating the gradient of its displacement-time
graph.
Example
Find Jane’s velocity when she is
a) going for the first bean bag.
b) returning from the first bean bag.
Solution
a) Velocity= gradient of displacement-time graph
= 20÷10
= 2m/s
b) Velocity= gradient
= -20 ÷10
= -2 m/s
The negative sign indicates Jane is now moving in the opposite direction.
Average speed and Average Velocity
Throughout the race Jane ran at a constant speed of 2m/s but her velocity varied
because the direction varied. In everyday speech the words speed and velocity are used
as if they mean the same thing. Mathematically they are different. Velocity is a vector
having direction and magnitude (size). Speed is a scalar having only magnitude.
Average speed and average velocity are also different.
Average speed = total distance travelled ÷ total time taken
Average Velocity = total displacement ÷ total time taken
Example
Find Jane’s average speed and average velocity in the bean bag race.
Solution
Average speed
Total distance = 20 + 20 + 30 + 30 + 30
= 130 m
Total time
= 65 s
Average speed = 130 ÷ 65
= 2m/s
As Jane travelled at a constant speed of 2m/s this is no surprise!!
Average velocity
When the race started Jane was at the basket. When the race finished she was 30m
from the basket in the negative direction.
Average velocity = -30 ÷ 65
= -0.462 m/s
So the average velocity is not the same as the average speed.
Example
A particle moves along a straight line so that its displacement , x metres, from a fixed
point O on the line at time t seconds is given by:
x= -3t for o≤t≤4 and x=2t-20 for 4≤t≤15
a) Find the displacement of the particle from O when:
(i) t=4
(ii) t=10
b) Sketch the (t,x) graph for the motion.
c) Find the velocity of the particle for
(i) 0≤t≤4
(ii) 4≤t≤15
d) Find the total distance travelled in the 15 seconds of motion.
e) Find the average velocity.
Solution
a) (i) When t=2, x=-3t
x = -3 x 2
x = -6m
(ii) When t=10, x=2t-20
x=2 x 10 -20
x=0
b) The equations x=-3t and x=2t-20. To draw the graphs it is useful to find the
start and end point of each segment.
0≤t≤4
x=-3t:
When t=0, x=0
When t=4, x=-12
4≤t≤15 x=2t-20: When t=4, x = 8-20
x= -12
When t=15,x= 30-20
x = 10
15
10
10
x (m)
5
0
-5 0
-10
0
4
15
-12
-15
t (s)
c) (i) In the first section: Velocity = gradient
= -12 ÷ 4
= -3 m/s
(ii) In the second section: Velocity = gradient
= 22 ÷ 11
= 2 m/s
d) Adding the distance travelled in each section gives:
Total distance travelled = 12 + 22 = 34m
e) Total displacement = 10m (displacement at t=15 is 10m)
Average velocity = total displacement ÷ total time taken
= 10 ÷ 15
= 0.667 m/s
Velocity-time graphs
A velocity-time graph always has time along the horizontal axis and velocity along
the vertical axis.
A velocity-time graph can be abbreviated to v-t graph or (t,v) graph.
Acceleration
Acceleration is the rate at which velocity is increasing.
Acceleration= increase in velocity ÷ time taken
Acceleration is given by the gradient of a velocity-time graph.
Displacement
The area between a velocity-time graph and the time axis gives the displacement.
Example 1
Positive
Direction
This sketch shows a dog chasing a ball. The dog accelerates to a velocity of 8m/s in
two seconds, runs at this velocity for 4 seconds and then slows down and stops to
pick up the ball. On the way back to its owner it accelerates to a speed of 6m/s and
reaches its owner after a total time of 18 seconds.
Velocity (m/s)
Here is the velocity-time graph for the dog’s motion.
10
8
6
4
2
0
-2 0
-4
-6
-8
A
B
C
2
6
8
G
D
9
10
E
Time (s)
17
F
18
What was the dog’s acceleration for each part of the motion?
Solution 1
Acceleration= gradient
OA: Acceleration=8 ÷ 2= 4 m/s²
BC: Acceleration = -8 ÷ 2 = -4m/s² (or decelerating at 4m/s²)
DE: Acceleration= -6 ÷ 1= -6m/s²
The acceleration here is negative although the dog is not slowing down. When the
acceleration and velocity are both negative the speed is decreasing but in a negative
direction.
FG: Acceleration: 6 ÷ 1 = 6 m/s²
The acceleration is in the opposite direction to the velocity and the dog is slowing down.
When the velocity and acceleration have the same sign the speed is increasing.
When the velocity and acceleration have opposite signs the speed is decreasing.
Example 2
8
Velocity (m/s)
6
4
2
0
-2 0
8
12
-4
Time (s)
Use the velocity-time graph to:
a) Find the initial velocity
b) Find the final velocity
c) Find the acceleration
d) Find the total displacement
Solution 2
a) Initial velocity = 6m/s
b) Final velocity = -3 m/s
c) Acceleration= gradient = -6 ÷ 8
e) Find the total distance travelled
f) Find the average velocity
g) Find the average speed
h) Sketch a speed-time graph.
= 0.75 m/s²
d) Displacement in first 8 seconds= ½ x 8 x 6
= 24 m
Displacement in last 4 seconds = -½ x 4 x 3
= -6m
Total displacement = 24 – 6 = 18m
Speed (m/s)
e) Total distance travelled = 24 + 6 = 30m
f) Average velocity = total displacement ÷ total time
= 18 ÷ 12 = 1.5 m/s
g) Average speed = total distance ÷ total time
= 30 ÷ 12
= 2.5 m/s
h)
7
6
5
4
3
2
1
0
0
8
12
Time (s)
Notice that the speed-time sketch is always positive.
Example 3
Velocity (m/s)
p
Q
u
R
0
2
18
24
Time (s)
The sketch shows the velocity-time graph for a fork lift truck as it transfers good
from one part of a warehouse to another. Find the value of u if the truck travels 60m.
Solution 3
Displacement is given by the area of the trapezium OPQR.
Area of trapezium OPQR= 60
½(16 + 24) u = 60
20u = 60
u= 3