c
(c) Let man will catch the bus after \('t'\) sec . So he will cover distance \(ut\).

Similarly distance travelled by the bus will be \(\frac{1}{2}a{t^2}\). For the given condition

\(u\;t = 45 + \frac{1}{2}a\;{t^2}\)\( = 45 + 1.25\;{t^2}\) \([As\;a = 2.5m/{s^2}]\)

\(⇒\)  \(u = \frac{{45}}{t} + 1.25\;t\)

To find the minimum value of \(u\)

\(\frac{{du}}{{dt}} = 0\) so we get \(t = 6\sec\) then,

\(u = \frac{{45}}{6} + 1.25 \times 6 = 7.5 + 7.5 = 15\,m/s\)