A man is 48 m behind a bus which is at rest. The bus starts accelerating at 1 m/$s^2$ at the same time the man starts running with uniform velocity 10 m/s. What is the minimum time taken by the man to reach the bus?

Let after time 't' man catches the bus.
After time 't' bus has covered the distance = $\frac{1}{2}\left(1\right)t^2=\frac{t^2}{2}$
Therefore the total distance man has to cover = $48+\frac{t^2}{2}$
Therefore, $vt=s$
$\Rightarrow 10t=48+\frac{t^2}{2}\Rightarrow 20t=96+t^2\Rightarrow t^2-20t+96=0\Rightarrow (t-12)(t-8)=0\Rightarrow t=12ort=8$
Therefore, minimum time in which man catches the bus is 8 seconds.