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LearnYMA01WME01 · Mechanics 1Kinematics of a particle in a straight lineDisplacement–time and velocity–time graphs
Kinematics of a particle in a straight line · Subtopic 3 of 3

Displacement–Time and Velocity–Time Graphs

On a velocity–time graph, the gradient is the acceleration and the area underneath is the displacement. On a displacement–time graph, the gradient is the velocity. Multi-stage journeys become trapezia, and an unknown time or top speed comes from setting the area equal to the total distance.

Edexcel YMA01 WME01≈ 35 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Sketch velocity–time graphs for multi-stage journeys.
  • Use gradient for acceleration and area for displacement.
  • Find unknowns (a time or a speed) from the area.
  • Interpret displacement–time graphs.
Pause and think

A train speeds up, cruises, then slows to a stop. Sketch its velocity against time. What shape is the region under the graph?

Section 2 of 14

Core idea

Start hereA journey in one picture

A train speeds up, cruises, then brakes. Draw speed against time and the whole story is in one shape: a trapezium. Its slopes are accelerations and its area is the distance travelled.

The big idea

On a velocity–time graph: gradient = acceleration, area = displacement.

1. A train journey

A train between two stations
204060801001205101520250t (s)v (m/s)distance = areaacceleratecruise at 20 m/sbrake
Area =12(60+130)(20)=1900=\tfrac12(60+130)(20)=1900 m. Acceleration =2030=0.67=\tfrac{20}{30}=0.67 m s−2^{-2}; deceleration =2040=0.5=\tfrac{20}{40}=0.5 m s−2^{-2}.
Why area is distance
  1. δs≈v δt\delta s\approx v\,\delta t
    Over a tiny time, speed is nearly constant.
  2. s=∑v δts=\sum v\,\delta t
    Thin strips under the graph.
The total is the area under the graph.
Worked exampleAccelerate, cruise, brake

30 s up to 20 m s−1^{-1}, steady, 40 s braking; total 130 s. Distance and accelerations?

Goal
Distance; rates.
Strategy
Trapezium area; gradients.
  1. 130−30−40=60 s130-30-40=60\text{ s}
    Steady stage.
  2. s=12(60+130)(20)=1900 ms=\tfrac12(60+130)(20)=1900\text{ m}
    Trapezium: parallel sides 60 and 130.
  3. 2030=0.67,−2040=−0.5\tfrac{20}{30}=0.67,\quad-\tfrac{20}{40}=-0.5
    Gradients.
Answer19001900 m; 0.670.67 and −0.5-0.5 m s−2^{-2}.
✓ CheckPieces: 300+1200+400=1900300+1200+400=1900 ✓.
Worked exampleUnknown top speed

Same times, total distance 2850 m. Find VV.

Goal
VV.
Strategy
Area in terms of VV.
  1. 12(60+130)V=95V\tfrac12(60+130)V=95V
    Area.
  2. 95V=2850⇒V=3095V=2850\Rightarrow V=30
    Solve.
Answer3030 m s−1^{-1}.
✓ Check1.5×1900=28501.5\times1900=2850 and 1.5×20=301.5\times20=30 ✓.
Exam tipSketch first Draw and label the graph before writing any equations.
Pause and think

What does a horizontal line on a velocity–time graph mean?

Summary

  • Gradient = acceleration.
  • Area = distance.
Key words
Velocity–time graph
Velocity plotted against time.
HistoryOresme's graph. Around 1350 the French scholar Nicole Oresme drew what was essentially a velocity–time graph and used the area of a triangle to show the "mean speed theorem": steady acceleration covers the same distance as moving at the average speed. That is s=12(u+v)ts=\tfrac12(u+v)t, three centuries before Galileo.
Fun factUsain Bolt's graph. Split-time data from Usain Bolt's 9.589.58 s world record show he reached top speed (over 1212 m s⁻¹) around the 6060–7070 m mark, then slowed very slightly. His velocity–time graph rises steeply, flattens, and dips a little at the end.
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