A car 2 m wide is parked along the edge of a road. At the instant, the car is started, a boy standing 6 m ahead of the car begins to cross the road with constant speed vs. If the acceleration of the car 1.5 ms^-2 and if it continues to move along the edge, then the minimum value of $v_0=(n{)}^{\frac{1}{2}}{\mathrm{ms}}^{-1}$ for which boy can avoid collide. The value of n is ____

$\text{ Here, }t=\frac{d}{v_0\sin \theta }=\frac{2}{v_0\sin \theta }$

$\begin{array}{l}\text{ and }6+v_{0}t\cos \theta =\frac{1}{2}a{t}^2\\ \Rightarrow 6+v_{0}t\cos \theta =\frac{1.5}{2}{t}^2\\ \Rightarrow 6+\left(v_0\cos \theta \right)\frac{2}{v_0\sin \theta }=\left(\frac{1.5}{2}\right){\left(\frac{2}{v_0\sin \theta }\right)}^2\\ \Rightarrow 6+2\cot \theta =\frac{3}{6+2\cot \theta }\\ \Rightarrow v_0^2{\sin }^2\theta =\frac{3}{6+2\cot \theta }\\ \Rightarrow v_0^2=\frac{3}{(6+2\cot \theta ){\sin }^2\theta }\end{array}$

For minimum value of $v_0^2,$

$y=(6+2\cot \theta ){\sin }^2\theta \text{ should be maximum. }$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{dy}{d\theta }=12\sin \theta \cos \theta +2\cos 2\theta \\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\sin 2\theta +2\cos 2\theta =0\\ \text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\tan 2\theta =-\frac{2}{6}=-\frac{1}{3}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{v}_{0min}^2=\frac{3}{6{\sin }^2\theta +\sin 2\theta }\end{array}$

$\mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} \theta ,v_{0min}^2=8\Rightarrow v_{0min}=\sqrt{8}{\mathrm{ms}}^{-1}$

$\therefore n=8$