Question

A bus started from rest with a uniform acceleration of 2m/s. A mate who is 9m behind the bus runs after it immediately with a uniform speed of 6m/s. calculate the time taken for the mate to draw level with the rear of the bus.

Answer 1

The bus's position $x$ at time $t$ is

$x_{\rm bus}=\dfrac12\left(2\,\dfrac{\rm m}{\rm s^2}\right)t^2$

If we take the bus's rear's starting position to be the origin, then the position of the mate (?) is given by

$x_{\rm mate}=-9\,\mathrm m+\left(6\,\dfrac{\rm m}{\rm s}\right)t$

The mate draws level with the rear of the bus when $x_{\rm bus}=x_{\rm mate}$:

$\dfrac12\left(2\,\dfrac{\rm m}{\rm s^2}\right)t^2=-9\,\mathrm m+\left(6\,\dfrac{\rm m}{\rm s}\right)t$

Drop the units to make things simpler:

$\dfrac22t^2=-9+6t\iff t^2-6t+9=(t-3)^2=0\implies t=3$

So it would take 3 seconds for the mate to catch up to the bus.