A man is at a distance of 6 ~m from a bus. The bus begins to move...

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 or zero. Minimum $v$ corresponds to $\frac{4 v^2}{9}-16=0$ which gives $v=6 \mathrm{~m} \mathrm{~s}^{-1}$.