The graph below shows the velocity - time graph of the motion throughout the journey i.e. from point $O$ to $C$irst, we find the time for the acceleration of the bus using $v=u+at$
when $t=t_1$,
$$\begin{array}{rl}& v=u+a{t}_1\\ & v=0+a{t}_1\\ & t_1=v/a\\ & t_1=\frac{9}{1.5}=6m/s^2\end{array}$$
second, we find the time for deceleration between points $B$ and $C$,
between these points
$u=9m/s,v=0$ (at point $C$ )
$$\begin{array}{rl}& 0=9+(-2\times t_3)\\ & -9=-2t_3,\\ & t_3=9/2=4.5s\end{array}$$
let $t_2$ be the time for the uniform velocity (points $A$ to $B$ ). using
total distance travelled $=$ area under the vel. $-$ time graph $s=$ area of a trapezium
$s=\frac{1}{2}(|AB|+|OC|)\times h$ or $s=\frac{1}{2}(t_2+|OC|)\times v$
where $|OC|=6+t_2+t_3$
$|OC|=6+t_2+4.5$
$|OC|=10.5+t_2$
and $h=v=9m/s$
$500=\frac{1}{2}(t_2+10.5+t_2)\times 9$
$\frac{500\times 2}{9}=2t_2+10.5$
$111.11-10.5=2t_2$
$t_2=\frac{100.61}{2}=50.3s$
$\therefore$ total time $=t_1+t_2+t_1$