# A bus starts from rest with an acceleration of $1\;ms^{-2}$ A man who is $48\;m$ behind the bus starts with a uniform velocity of $10\;ms^{-1}$. The minimum time after which the man will catch the bus is

$\begin{array}{1 1} 4.8\;s \\ 8\;s \\ 10\;s \\12\;s \end{array}$

Answer : $8\;s$
$10t =48+\large\frac{1}{2}$$+ 1 \times t^2$
or $t^2-20t+96=0$
or $t^2-8t-12t+96=0$
or $t(t-8) -12 (t-8) -12 (t-8)=0$
or $(t-12) (t-8) =0$
or $t=8s$ or $12\;s$
But we are interested in minimum time.
Your answer is correct, no doubt in it but write the exact statement  before solving any problem.  You haven't explained  the frame of reference  between man and the bus.
Good Luck.