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 \(v=u+a t\)
when \(t=t_{1}\),
$$
\begin{aligned}
&v=u+a t_{1} \\
&v=0+a t_{1} \\
&t_{1}=v / a \\
&t_{1}=\frac{9}{1.5}=6 m / s ^{2}
\end{aligned}
$$
second, we find the time for deceleration between points \(B\) and \(C\),
between these points
\(u=9 m / s , v=0\) (at point \(C\) )
$$
\begin{aligned}
&0=9+\left(-2 \times t_{3}\right) \\
&-9=-2 t _{3}, \\
&t_{3}=9 / 2=4.5 s
\end{aligned}
$$
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
\(s=\frac{1}{2}(|A B|+|O C|) \times h\) or \(s=\frac{1}{2}\left(t_{2}+|O C|\right) \times v\)
where \(|O C|=6+t_{2}+t_{3}\)
\(|O C|=6+t_{2}+4.5\)
\(|O C|=10.5+t_{2}\)
and \(h=v=9 m / s\)
\(500=\frac{1}{2}\left(t_{2}+10.5+t_{2}\right) \times 9\)
\(\frac{500 \times 2}{9}=2 t_{2}+10.5\)
\(111.11-10.5=2 t_{2}\)
\(t_{2}=\frac{100.61}{2}=50.3 s\)
\(\therefore\) total time \(=t_{1}+t_{2}+t_{1}\)