(c)

If the man did not run, the bus would be at a distance $s_1$ at time $t$ given by

$s_1=6+\frac{1}{2} a t^2=6+\frac{1}{2} \times 3 \times t^2=6+\frac{3}{2} t^2$

If $v$ is the speed of man, he would cover a distance $s_2=v t$ in time $t$. To catch the bus, $s_1=s_2$

$6+\frac{3}{2} t^2=v t \quad$ or $\quad t^2-\frac{2 v}{3} t+4=0$

which gives $t =\frac{2 v}{6} \pm \frac{1}{2}\left[\frac{4 v^2}{9}-16\right]^{1 / 2}$

Now, t will be real if $\left(\frac{4 v^2}{9}-16\right)$ is positive orzero. Minimum $v$ corresponds to $\frac{4 v^2}{9}-16=0$ which gives $v=$

$6\,m s^{-1}$.