A bus starts to move with an acceleration of $1m/s^2$. A boy who is $48m$ behind the bus starts running at $10m/s$. The minimum time (in $s$) at which the boy can catch the bus is

Given,
Acceleration of bus $\left(a\right)=1m/s^2$
Distance between boy and bus $\left(D\right)=48m$
velocity of bus $\left(v\right)=10m/s$
Suppose man will catch the bus after time $t$.
Since the distance from man position is same for both, then
Distance covered by bus
$=$ Distance covered by man$-48m$

$⟹\frac{1}{2}a{t}^2=vt-48$

$⟹\frac{1}{2}\left(1\right)t^2=10t-48$

$⟹t^2-20t+96=0$

$⟹(t-12)(t-8)=0$

$⟹t=8s$ (Minimum Value)

or $t=12s$

So,

at $t=8s$, the man will catch the bus
Final answer: 8$s$