Lesson 4 of 6
Speed-time graphs
I can plot and interpret distance-time and speed-time graphs for motion in one direction.Syllabus
Syllabus K223 / K224, 2(e). Plot and interpret distance-time and speed-time graphs for motion in one direction.
I can use speed-time shapes to identify rest, uniform speed, uniform and non-uniform acceleration.Syllabus
Syllabus K223 / K224, 2(g). Use speed-time shapes to identify rest, uniform speed, uniform and non-uniform acceleration.
I can calculate distance from area for uniform speed or acceleration in one direction.Syllabus
Syllabus K223 / K224, 2(h). Calculate distance from area for uniform speed or acceleration in one direction.
In short
- The gradient of a speed-time graph is the acceleration.
- The area under the line is the distance travelled.
- A horizontal line means constant speed.
Height gives speed. Gradient gives acceleration. Area gives distance.
Here a gradient measures change in velocity per second. It has a different meaning from the gradient of a distance-time graph because the vertical quantity has changed.
The vertical axis now shows speed at each time. A horizontal line on this graph means constant speed. The object is at rest only if that line is at zero.
| Graph shape | Motion |
|---|---|
| Line at zero | At rest throughout that interval. |
| Horizontal line above zero | Constant speed; zero acceleration. |
| Straight sloping line | Uniform acceleration. Its gradient is constant. |
| Curve | Non-uniform acceleration. Its gradient changes. |
A curve becoming less steep still shows a changing speed, but the magnitude of acceleration is getting smaller. For a sloping straight segment, calculate acceleration using the change in speed divided by the elapsed time. In the one-direction journey below, forwards is positive.
A curve means changing acceleration
This example moves forwards throughout. Its graph becomes steeper: the speed increases by a larger amount each second.
| Time / s | Speed / (m/s) |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |
| 3 | 9 |
Over the successive one-second intervals, the changes are 1, 3 and 5 m/s. The average accelerations over those intervals are therefore 1, 3 and 5 m/s2. The acceleration is non-uniform. A straight rising line would instead have equal changes in equal times.
Why area gives distance
At constant speed, distance = speed x time. On this graph, the height is speed and the width is time, so their product is the area of a rectangle. When speed changes uniformly, use the area under the sloping line, splitting it into rectangles and triangles where needed.
Motion and graphs
Moving in one direction
A trolley moves along a straight track in one direction. It travels 18 m in 6 s.
Move the slider to see the trolley's position and the values on both graphs at the selected time. You can also use the arrow keys.
Moving forwards at a constant 4 m/s.
- Distance
- 6 m
- Speed
- 4 m/s
- Acceleration
- 0 m/s2
View the data table
| Time / s | Distance / m | Speed / (m/s) |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 1 | 2 |
| 2 | 4 | 4 |
| 3 | 8 | 4 |
| 4 | 12 | 4 |
| 5 | 16 | 4 |
| 6 | 18 | 0 |
Worked example
Find the distance and average speed
The trolley reaches 4 m/s in 2 s, stays at 4 m/s for 3 s, then brakes to rest in 1 s. Use the one-direction example above. Split the area under the graph into two triangles and a rectangle.
- 0 to 2 s, triangle: (1/2) x 2 s x 4 m/s = 4 m.
- 2 to 5 s, rectangle: 3 s x 4 m/s = 12 m.
- 5 to 6 s, triangle: (1/2) x 1 s x 4 m/s = 2 m.
Total distance = 4 + 12 + 2 = 18 m
Average speed = 18 / 6 = 3 m/s
Multiplying time by speed gives metres: s x m/s = m. The area counts distance throughout the one-direction journey.
Find the accelerations from the gradients
First interval: (4 - 0) / 2 = 2 m/s2.
Middle interval: (4 - 4) / 3 = 0 m/s2.
Final interval: (0 - 4) / 1 = -4 m/s2.
The trolley loses 4 m/s of velocity each second while braking, compared with a gain of 2 m/s each second at the start.
Try it step by step
Work it out in steps: Area of the triangle in m, then Area of the rectangle in m, then Total distance in m.
Show the answer
Problem 1.
Area of the triangle in m: Triangle: 1/2 x 4 x 6 = 12 m
Area of the rectangle in m: Rectangle: 5 x 6 = 30 m
Total distance in m: Distance = 12 + 30 = 42 m
Problem 2.
Area of the triangle in m: Triangle: 1/2 x 5 x 8 = 20 m
Area of the rectangle in m: Rectangle: 10 x 8 = 80 m
Total distance in m: Distance = 20 + 80 = 100 m
Problem 3.
Area of the triangle in m: Triangle: 1/2 x 6 x 12 = 36 m
Area of the rectangle in m: Rectangle: 4 x 12 = 48 m
Total distance in m: Distance = 36 + 48 = 84 m
Check the vertical axis. Area under a distance-time graph does not give distance. The area rule here works because the vertical axis is speed.
Check your understanding
On a speed-time graph, the line is horizontal at 3 m/s for 4 s. What happens?
Show the answer
Acceleration is zero and distance travelled is 12 m.
The gradient is zero. The rectangular area is 3 m/s x 4 s = 12 m. The object keeps moving throughout.
Practise an exam question
Write your answer first, then tick the mark points it makes.
Exam-style question
A cyclist starts from rest and accelerates uniformly to 6.0 m/s in 4.0 s. She then rides at a constant 6.0 m/s for a further 10 s. Calculate the total distance she travels in the 14 s. [2]
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