The big idea: Acceleration tells you how quickly the velocity is changing.
Speeding up, slowing down, or changing direction — they all count as acceleration.
Unit: m s⁻² (metres per second, each second).
Spot it on the graph: Line going up = speeding up · line flat = steady (no acceleration) · line going down = slowing down.
On a velocity–time graph the slope of the line is the acceleration: how much the velocity changes ÷ the time it takes.
[Diagram: phys-formula-triangle] - Available in full study mode
- final velocity (m s⁻¹)
- initial velocity (m s⁻¹)
- acceleration (m s⁻²)
- time (s)
What does 'suvat' mean?: suvat is just the five letters in any constant-acceleration problem. You're usually given three of them and use an equation (like v = u + at) to find a fourth:
| Quantity | What it means | |
|---|---|---|
| s | displacement | how far it goes |
| u | initial velocity | speed at the start |
| v | final velocity | speed at the end |
| a | acceleration | how fast the velocity changes |
| t | time | how long it takes |
Worked example — find the acceleration
A train speeds up from 8.0 m s⁻¹ to 20 m s⁻¹ in 6.0 s. Find its acceleration.
Solution
- Start with the given formula:
- Put in the numbers (v = 20, u = 8.0, t = 6.0):
- Rearrange and solve — keep the unit:
Final answer
a = 2.0 m s⁻².
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How this is tested: Acceleration is rarely tested on its own — it lives inside suvat and motion-graph questions.
- Paper 1A: read it off a graph — the slope of a velocity–time graph, or the area under an acceleration–time graph. - Paper 2: you use a in the suvat equations (the constant-acceleration formulas like v = u + at).
Watch the sign: slowing down gives a negative acceleration.
Acceleration–time graphs: On an acceleration–time (a–t) graph the area under the line = the change in velocity. Starting from rest, that area is the velocity reached.
IB-style question — velocity from an a–t graph
An object starts from rest. On its acceleration–time graph the acceleration is constant at 3.0 m s⁻² for 5.0 s. Find its velocity at 5.0 s.
Solution
- Start with the given formula:
- It starts from rest (u = 0), so put in the numbers:
- Work it out — this equals the area under the a–t graph:
Final answer
v = 15 m s⁻¹ — the same as the area under the a–t graph (3.0 × 5.0).