1.1.5Forces and Motion
Velocity-Time Graphs — Edexcel International GCSE Science Double Award 4SD0 - Physics Revision Notes & Topical Questions
5 min read 8 parts Lesson 6 of 17
Part 1 of 6What the graph shows
A velocity-time graph plots velocity (m/s) on the vertical axis against time (s) on the horizontal axis.
Part 2 of 6Reading the shape
- For a falling object with air resistance, the curve flattens and becomes horizontal at the terminal velocity.
Be carefula velocity-time graph that is horizontal does not mean the object is stationary, unless the line is at zero.
Part 3 of 6Gradient equals acceleration
The gradient of a velocity-time graph is the change in velocity divided by the change in time, which is exactly the definition of acceleration:
acceleration = gradient = Δv ÷ Δt
For a straight-line section, draw a right-angled triangle against the line, read off the vertical change (Δv, in m/s) and the horizontal change (Δt, in s) and divide. The unit is m/s². For a curve, the acceleration at a moment is the gradient of the tangent at that point.
A downwards-sloping line gives a negative gradient and therefore a negative acceleration (deceleration).
Part 4 of 6Describing a whole journey
Most journeys are split into sections.
- and (c) a downwards sloping line as it brakes to a stop.
- and how long it lasted.
Remember also that distance travelled is the area under the graph; the next lesson shows you how to work this out.
Part 5 of 6Drawing a velocity-time graph
To draw a graph from a description, first decide on scales for the axes, so that the data fills most of the grid.
- and curve to flat in the last.
- Being able to convert between the two types of graph is a good test of understanding.
- Look at 1.1.1 and 1.1.6 for the matching ideas.
- and where the lines cross the two objects have the same velocity at that instant, though not necessarily the same position.
Part 6 of 6Worked examples
A velocity-time graph is a straight line from (0 s, 0 m/s) to (6.0 s, 15 m/s). Find the acceleration.
Gradient = Δv ÷ Δt = 15 ÷ 6.0 = 2.5 m/s².
A cyclist is moving at 10 m/s. The graph slopes down in a straight line to 0 m/s after 8.0 s. Find the acceleration.
Gradient = (0 − 10) ÷ 8.0 = −1.25 m/s², so the deceleration is 1.25 m/s².
A graph has three sections: 0–4 s a line rising from 0 to 8 m/s; 4–10 s horizontal at 8 m/s; 10–12 s falling from 8 m/s to 0. Find the acceleration in each section.
0–4 s: (8 − 0) ÷ 4 = 2.0 m/s².
4–10 s: 0 m/s².
10–12 s: (0 − 8) ÷ 2 = −4.0 m/s². The stopping is twice as fast as the start, so the line is twice as steep.
Key terms
| Term | Meaning |
|---|---|
| Velocity-time graph | A graph showing how velocity varies with time. |
| Gradient | Δv ÷ Δt = acceleration. |
| Uniform acceleration | Constant acceleration; a straight sloping line. |
| Terminal velocity | Constant velocity when forces balance; a horizontal line at the end of a curve. |
| Deceleration | Negative acceleration; a downward-sloping line. |
Avoid these mistakes
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