What are the key learning points about distance-time graphs?
A horizontal line means the object is not moving.
The gradient of the line = the speed.
The steeper the line, the faster the object is moving.
What do distance-time graphs show?
A horizontal line on a A graph with distance travelled plotted on the vertical axis against time taken on the horizontal axis. shows that the object is Not moving, stopped, at rest (speed = 0 m/s). (not moving because the distance does not change).
A sloping line on a distance-time graph shows that the object is moving.
Key fact
In a distance-time graph, the slope or In a graph, the gradient is the steepness of the line. The greater the gradient, the greater the rate of change. of the line is equal to the speed of the object.
The steeper the line (and the greater the gradient) the faster the object is moving.
How much do you know about distance-time graphs?
See how much you know about distance-time graphs by answering the questions below.
Question
Calculate the speed of the object represented by the green line in the graph, from 0 to 3 s.
Calculate the average speed of the object represented by the green line in the graph, from 0 to 10 s.
Calculate the average speed of the object represented by the purple line in the graph, from 0 to 2 s.
Answer
Distance moved = (6 m – 0 m) = 6 m
time taken = (3 s – 0 s) = 3 s
speed = gradient of distance-time graph = distance moved ÷ time taken
speed = 6 m ÷ 3 s = 2 m/s
The total distance travelled by the vehicle on the green line is 7 m.
The time taken to travel this distance is 10 s.
Average speed = total distance moved ÷ time taken = 7 m ÷ 10 s = 0.7 m/s.
The speed of the object represented by the purple line can also be calculated.
distance moved = (10 m – 0 m) = 10 m
time taken = (2 s – 0 s) = 2 s
speed = gradient of distance-time graph = distance moved ÷ time taken
speed = 10 m ÷ 2 s = 5 m/s
Question
Look at this distance-time graph and answer the following questions.
Question
How far did the vehicle travel in the first 4 seconds?
Answer
30 m
Question
What was the speed of the vehicle over the first 4 seconds?
Answer
distance moved = (30 m – 0 m) = 30 m
time taken = (4 s – 0 s) = 4 s
speed = gradient of distance-time graph = distance moved ÷ time taken
speed = 30 m ÷ 4 s = 7.5 m/s
Question
How long was the vehicle stationary?
Answer
The distance moved remains at 30 m between 4 and 8 seconds. The vehicle was stationary for a total of 4 seconds.
Question
What was the average speed of the vehicle over the journey?
Answer
distance moved = (40 m – 0 m) = 40 m
time taken = (10 s – 0 s) = 10 s
average speed = total distance moved ÷ time taken = 40 m ÷ 10 s = 4 m/s
How is acceleration shown on distance-time graphs? (Higher tier only)
If the speed of an object changes, it will be The rate of change of velocity is measured in metres per second squared. Acceleration = change of velocity ÷ time taken. or decelerating.
This can be shown as a curved line on a distance-time graph.
The table shows what each section of the graph represents:
| Section of graph | Gradient | Speed |
|---|---|---|
| A | Increasing | Increasing |
| B | Constant | Constant |
| C | Decreasing | Decreasing |
| D | Zero | Stationary (at rest) |
What do displacement-time graphs show? (Higher tier only)
Displacement-time graphs show how the The total number of metres (m) moved in a specific direction. Displacement is a vector quantity. of a moving object changes with time.
A horizontal line on a displacement-time graph shows that the object is stationary (not moving because the displacement does not change)
A sloping line on a displacement-time graph shows that the object is moving.
In a displacement-time graph, the slope or gradient of the line, is equal to the The rate of change of displacement. The distance travelled in one second in a specified direction. Measured in m/s. of the object.
The steeper the line (and the greater the gradient) the faster the object is moving.
Key fact
- The gradient of a displacement–time graph = velocity.
Example
| Section of graph | What is represented |
|---|---|
| O - A | Positive constant velocity |
| velocity = gradient of displacement-time graph = displacement ÷ time | |
| displacement = 4 m – 0 m = 4 m | |
| time = 2 s – 0 s = 2 s | |
| velocity = displacement ÷ time = 4 m ÷ 2 s = 2 m/s | |
| A - B | Stopped |
| the displacement does not change: velocity = 0 m/s | |
| B - C | Negative constant velocity |
| the object is now travelling in the opposite direction | |
| velocity = gradient of displacement-time graph = displacement ÷ time | |
| displacement = 0 m – 4 m = -4 m | |
| time = 8 s – 7 s = 1 s | |
| velocity = displacement ÷ time = -4 m ÷ 1 s = -4 m/s | |
| At C | the object has returned to its starting position |
| the displacement is once again zero |
How much do you know about displacement-time graphs?
Carol is jogging at constant velocity.
Below is the displacement-time graph for the first 10 s of her jog.
Question
What is Carol’s displacement after 6 s?
Answer
Carol’s displacement is 24 m.
Question
What is Carol’s velocity?
Answer
velocity = gradient of displacement-time graph = displacement ÷ time
displacement = 40 m – 0 m = 40 m
time = 10 s – 0 s = 10 s
velocity = displacement ÷ time= 40 m ÷ 10 s = 4 m/s
Carol’s velocity is 4 m/s.
Question
If Carol continues to jog at the same velocity how far will she have travelled after 22 seconds?
Answer
displacement = velocity x time
displacement = 4 m/s x 22 s = 88 m
Carol will have travelled 88 m.
See how much you know about displacement-time graphs by taking the quiz below.
More on Unit 1: Motion
Find out more by working through a topic
- count4 of 4
- count1 of 4
- count2 of 4
