velocity-time graph of car police example

The v-t curve for the car is represented by the red line, while that v-t curve for the police car is the blue line.

We need to find the unknown time t (in seconds), when the police catch up to the car. We find this by comparing the distance travelled by each (it will be the same distance at the overtaking point.) So we need to set the area under the v-t curve for the car (the pink shaded area) to be equal to the area under the v-t curve for the police car (the blue shaded area).

a. The area under the curve for the car at time t (in seconds) is simply `72t`. (It is a rectangle, `72` high and width t).

The area under the trapezium (trapezoid) for the police car at unknown time t, using `A = (a + b)h/2` is:

`(t + t - 10) × 90/2 = 45 (2t - 10)`

We set these equal to find the required time: `72t = 45 (2t - 10)`

That is, when `72t = 90t - 450`

So `t = 25\ "s"` will be the time the police car catches up.


b. Both of the cars have travelled `72 × (25/3600) × 1000 = 500\ "m"` during those `25\ "s"`.

[We have used `d = s × t` and converted from seconds to hours (since the velocities are given in `"km/h"`).]