Area under the line
A velocity-time line has a shape under it. The area of that shape is the displacement of the object.
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- Velocity climbs from 2 m/s to 10 m/s in 4 s.
- Shade the shape between the line and the time axis.
- Its heights are 2 m/s and 10 m/s. Its width is 4 s.
- Average the heights, times the width: 24 m.
The rule
The area under a velocity-time line is the displacement. For constant acceleration, v = u + at and s = ut + ½at².
Here u is the first velocity and v is the last. The letters a, t and s are acceleration, time and displacement. In at, the letters a and t are multiplied. So are u and t in ut.
In the example, u = 2 m/s and v = 10 m/s. Here t = 4 s. So a = (10 − 2) ÷ 4 = 2 m/s².
The shape here is a trapezium. It has four sides, two of them parallel. The average height is (2 + 10) ÷ 2 = 6 m/s. Area = 6 × 4 = 24 m.
The equation gives the same. s = 2 × 4 + ½ × 2 × 4² = 8 + 16 = 24 m.
A third equation leaves out the time: v² = u² + 2as. Check: v² = 10² = 100. And u² + 2as = 4 + 2 × 2 × 24 = 4 + 96 = 100.
These equations work only when the acceleration is constant.
Braking distance
A car brakes with a = −5 m/s² until it stops. Then v = 0. Put v = 0 and a = −5 in v² = u² + 2as. Then 0 = u² + 2 × (−5) × s = u² − 10s. So s = u² ÷ 10.
Reaction time is how long a driver takes to start braking. The car keeps moving in that time. Braking distance does not include the reaction time. It is measured from the moment the brakes go on.
| Speed when braking starts | Braking distance |
|---|---|
| 10 m/s | 100 ÷ 10 = 10 m |
| 20 m/s | 400 ÷ 10 = 40 m |
Here 10² = 100 and 20² = 400.
Watch out!
Double the speed does not double the braking distance. Doubling the speed makes the distance four times as long. Because s = u² ÷ 10, doubling u makes u² four times as big.