91Ó°ÊÓ

Distance of the person from the bus, $s=9.0m$

The bus has an acceleration of $a=1m/s^2$

Initial speed of the person, $u=4.5m/s$

The relative acceleration between the bus and the person is, $a\text{'}=-1m/s^2$

Now, according to the question, the distance between the door and the person is given by,

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ 9=4.5t-\frac{1}{2}{t}^2 \\ 0.5t^2-4.5t+9=0 \\ {t}^2-9t+18=0 \\ {t}^2-6t-3t+18=0 \\ \left(t-3\right)\left(t-6\right)=0 \\ t=3s,t=6s \end{gathered}$$

It will take$3s$by the person to reach the bus and jump in.

Distance travelled by the person can be written as,

$$\begin{gathered} x=vt \\ x=4.5t \end{gathered}$$

Let $t\text{'}$be the time the person has to wait.

$d\text{'}$be the relative distance

$$\begin{gathered} 4.5t=d\text{'}+ut+0.5a{t}^2 \\ 4.5t=\left(9+0.5at{\text{'}}^2\right)+at\text{'}+0.5a{t}^2 \\ 4.5t=9+0.5t{\text{'}}^2+t\text{'}t+0.5t^2 \\ 0.5t^2-\left(4.5-t\text{'}\right)t+\left(9+0.5t{\text{'}}^2\right)=0 \end{gathered}$$

Thus we can obtain a solution of the quadratic equation as,

role="math" localid="1648494687212" $t=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$and$b^2-4ac>0$

Consider, $b=4.5-t\text{'}$and $4ac=18+t{\text{'}}^2$

Now,

$$\begin{gathered} b^2-4ac={\left(9-t\text{'}\right)}^2-18-t{\text{'}}^2 \\ {b}^2-4ac=81+t{\text{'}}^2-18t\text{'}-18-t{\text{'}}^2 \\ {b}^2-4ac=63-18t\text{'}>0 \\ 18t\text{'}<63 \\ t\text{'}=3.5s \end{gathered}$$