The big idea: On a velocity–time (v–t) graph, the area under the line tells you the displacement — how far the object has gone.
The slope told you the acceleration (last micro). The area tells you the distance.
Unit of that area: m s⁻¹ × s = m (metres) — exactly a displacement.
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).
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.
Worked example — displacement from rest
A car starts from rest and speeds up steadily to 20 m s⁻¹ in 5.0 s. Find its displacement.
Solution
- Start with the given formula (it starts from rest, so u = 0):
- Put in the numbers (u = 0, v = 20, t = 5.0):
- Work it out — keep the unit:
Final answer
s = 50 m — the same as the triangle's area (½ × 5.0 × 20).
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How this is tested: Finding displacement from a v–t graph is a classic graph skill.
- Paper 1A: an MCQ — pick the displacement, or compare the areas under two graphs. - Paper 1B / Paper 2: 'find the distance/displacement' — you read values off the graph and work out the area, often by splitting it into a rectangle and a triangle.
Classic trap: when the line dips below the time axis, that area is negative — the object is moving backwards, so it counts the other way.
Splitting the area: If the area isn't a single neat shape, chop it into a rectangle and a triangle, work out each, then add them. A trapezium = rectangle (height u) + triangle on top.
IB-style question — displacement of a train
A train is already moving at 6.0 m s⁻¹. It then speeds up steadily to 18 m s⁻¹ over the next 8.0 s. Find the distance it travels during these 8.0 s.
Solution
- The v–t line is a straight slope from u to v, so use the given trapezium area:
- Put in the numbers (u = 6.0, v = 18, t = 8.0):
- Work out the average velocity, then multiply by the time:
Final answer
s = 96 m — the area of the trapezium under the line (a 6.0-high rectangle plus a triangle on top).