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,

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