Distance-time graphs
Calculations involving speed, distance and time
The distance travelled by an object moving at constant speed can be calculated using the equation:
distance travelled = speed × time
\(\text{s} = \text{vt}\)
This is when:
- distance travelled (\(\text{s}\)) is measured in metres (m)
- speed (\(\text{v}\)) is measured in metres per second (m/s)
- time (\(\text{t}\)) is measured in seconds (s)
Example
A car travels 500 m in 50 s, then 1,500 m in 75 s. Calculate its average speed for the whole journey.
First calculate total distance travelled (\(\text{s}\)):
500 + 1,500 = 2,000 m
Then calculate total time taken, \(\text{t}\):
50 + 75 = 125 s
Then rearrange \(\text{s} = \text{vt}\) to find \(\text{v}\):
\(\text{v} = \frac{\text{s}}{\text{t}}\)
\(\text{t}\) = 2,000 ÷ 125
\(\text{t}\) = 16 m/s
If an object moves along a straight line, the distance travelled can be represented by a distance-time graph.
Example
Calculate the speed of the object represented by the green line in the graph, from 0 to 4 s.
change in distance = (8 – 0) = 8 m
change in time = (4 – 0) = 4 s
\(\text{speed} = \frac{\text{distance}}{\text{time}}\)
speed = 8 ÷ 4
speed = 2 m/s
Question
Calculate the speed of the object represented by the purple line in the graph.
change in distance = (10 – 0) = 10 m
change in time = (2 – 0) = 2 s
\(\text{speed} = \frac{\text{distance}}{\text{time}}\)
speed = 20 ÷ 2
speed = 5 m/s
Average speed
Many journeys do not occur at a constant speed. Bodies can speed up and slow down along the journey. However the average speed can still be found for a journey by:
Average speed = total distance travelled ÷ time
Example
Calculate the average speed of the entire journey of the object following the green line on the graph, from 0 s to 7 s.
Average speed = distance ÷ time
Average speed = 8 ÷ 7
Average speed = 1.14 m/s
Distance-time graphs for accelerating objects – Higher
If the speed of an object changes, it will be The rate of change in speed (or velocity) is measured in metres per second squared. Acceleration = change of velocity ÷ time taken. or Slowing down or negative acceleration, eg the car slowed down with a deceleration of 2 ms⁻².. 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) |
| Section of graph | A |
|---|---|
| Gradient | Increasing |
| Speed | Increasing |
| Section of graph | B |
|---|---|
| Gradient | Constant |
| Speed | Constant |
| Section of graph | C |
|---|---|
| Gradient | Decreasing |
| Speed | Decreasing |
| Section of graph | D |
|---|---|
| Gradient | Zero |
| Speed | Stationary (at rest) |
If an object is accelerating or decelerating, its speed can be calculated at any particular time by:
- drawing a A straight line that just touches a point on a curve. A tangent to a circle is perpendicular to the radius which meets the tangent. to the curve at that time
- measuring the gradient of the tangent
As the diagram shows, after drawing the tangent, work out the change in distance (A) and the change in time (B).
\(\text{Gradient} = \frac{\text{vertical change (A)}}{\text{horizontal change (B)}}\)
Note that an object moving at a constant speed is changing direction continually. Since The speed of an object in a particular direction. has an associated direction, these objects are also continually changing velocity, and so are accelerating.
More guides on this topic
- Newton's laws - AQA Synergy
- Circuits - AQA Synergy
- Mains electricity - AQA Synergy
- Acids and alkalis - AQA Synergy
- Rates of reaction - AQA Synergy
- Energy, rates and reactions - AQA Synergy
- Equilibria - AQA Synergy
- Electrons and chemical reactions - AQA Synergy
- Sample exam questions - movement and interactions - AQA Synergy
