Section 1 of 8What the graph shows
A velocity-time graph plots velocity (m/s) on the vertical axis against time (s) on the horizontal axis.
Section 2 of 8Reading 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.
Section 3 of 8Gradient 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).
Section 4 of 8Describing 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.
Section 5 of 8Drawing 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.
Section 6 of 8Worked 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.
Section 7 of 8Key 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. |
Section 8 of 8Exam technique
Marked practiceQuizzes for this lesson
Constant Acceleration Equations (SUVAT) & Slopes
v = u + at | s = ut + ½at2 | v2 = u2 + 2as | s = ½(u + v)t
- Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
- Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
- Free Fall in vacuum:
a = g ≈ 9.81 m/s2downwards (air resistance negligible).