Topic 4 of 6
Velocity-time graphs
Height gives velocity. Gradient gives acceleration. Area gives displacement.
Here a gradient measures change in velocity per second. It has a different meaning from the gradient of a displacement-time graph because the vertical quantity has changed.
The vertical axis now shows velocity at each time. A horizontal line on this graph means constant velocity. 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 positive velocity; 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 velocity, but the magnitude of acceleration is getting smaller. For a sloping straight segment, calculate acceleration using the change in velocity 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 velocity increases by a larger amount each second.
| Time / s | Velocity / (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 displacement
At constant velocity, displacement = velocity x time. On this graph, the height is velocity and the width is time, so their product is the area of a rectangle. When velocity changes uniformly, use the area under the sloping line, splitting it into rectangles and triangles where needed.
For displacement, count area above the time axis as positive and area below it as negative. For total distance, add the sizes of all the areas without cancelling them.
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.
- Displacement
- 6 m
- Velocity
- 4 m/s
- Acceleration
- 0 m/s2
View the data table
| Time / s | Displacement / m | Velocity / (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. All velocities here are positive, so displacement is also +18 m.
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.
Pure / Changing direction
Distance and displacement when direction changes
Select Changing direction above to follow this example. Velocity falls uniformly from +2 to -2 m/s over 4 s. The trolley slows to rest at 2 s, turns, then speeds up backwards.
- Acceleration: (-2 - 2) / 4 = -1 m/s2 throughout.
- Signed area before 2 s: (1/2) x 2 x 2 = +2 m.
- Signed area after 2 s: -(1/2) x 2 x 2 = -2 m.
- Displacement: +2 + (-2) = 0 m. The trolley returns to its start.
- Total distance: 2 + 2 = 4 m. Add the magnitudes of both areas.
Average speed is 4 / 4 = 1 m/s; average velocity is 0 / 4 = 0 m/s. At the turning instant, velocity is zero but acceleration is still -1 m/s2. There is no pause lasting a whole interval.
Check the vertical axis. Area under a displacement-time graph does not give displacement. The area rule here works because the vertical axis is velocity.