Explanation

(c) Let man will catch the bus after ‘$t$’$\sec$. So he will cover distance $ut$
Similarly distance travelled by the bus will be $\frac{1}{2}at^{2}$
For the given condition
$u\ t = 45 + \frac{1}{2}a\ t^{2} = 45 + 1.25t^{2}\ \ \ \lbrack As\ a = 2.5\ m/s^{2}\rbrack$
$\Rightarrow u = \frac{45}{t} + 1.25t$
To find the minimum value of $u$
$\frac{du}{dt} = 0$ sp we get $t = 6\ sec$ then,
$u = \frac{45}{6} + 1.25 \times 6 = 7.5 + 7.5 = 15\ m/s$