The big idea: Watch a car's speed on a velocity–time graph. The space under the line is how far it has travelled — its displacement.
(Last micro the slope gave the acceleration; now the area gives the distance.)
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Spot it on the graph: Slope of a v–t line = acceleration · area under a v–t line = displacement.
Flat line → the area is a rectangle. Sloping line → a triangle (or a trapezium).
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Split the area under the line into shapes you can do: a rectangle (length × width) and a triangle (½ × base × height).
For a single straight line from u to v over time t, there's a shortcut given in the data booklet — the area of the trapezium:
- displacement (m)
- initial velocity (m s⁻¹)
- final velocity (m s⁻¹)
- time (s)
Why it works: ½(u + v) is just the average velocity — halfway between the start velocity u and the end velocity v.
Average velocity × time = displacement. That is the area under the line.
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A car starts from rest and speeds up steadily to 20 m s⁻¹ in 5.0 s. Find its displacement.
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How this is tested — you find the distance from a v–t graph, which is the area under the line. Two ways it comes up:
Paper 1A — multiple choice
- Pick the displacement from the options.
- It's just the area under the line.
Paper 1B / Paper 2 — written
- Find the distance — read the graph.
- Work out the area (split into a rectangle + triangle if needed).
Watch out — area below the axis: When the line dips below the time axis, that area is negative — the object moves backwards. Subtract it for the net displacement.
Splitting the area: Not a neat shape? Split it into a rectangle + a triangle, find each, then add.
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A train is already moving at 6.0 m s⁻¹, then speeds up steadily to 18 m s⁻¹ over the next 8.0 s.
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