A 2-m wide truck is moving with a uniform speed $v_0=8{\mathrm{ms}}^{-1}$ along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed v when the truck is 4 m away from him. The minimum value of y so that he can cross the road safely is 

Let the man start crossing the road at an angle $\theta$ with the roadside. For safe crossing, the condition is that the man must cross the road by the time the truck describes the distance (4 + 2 cot $\theta$)

$\begin{array}{l}\text{ So, }\frac{4+2\cot \theta }{8}=\frac{2/\sin \theta }{v}\text{ or }v=\frac{8}{2\sin \theta +\cos \theta }\\ \text{ For minimum }v,\frac{dv}{d\theta }=0\end{array}$

$\begin{array}{r}\begin{array}{l}\mathrm{or} \frac{-8(2\cos \theta -\sin \theta )}{(2\sin \theta +\cos \theta {)}^2}=0\text{ or }2\cos \theta -\sin \theta =0\\ \text{ or }\tan \theta =2,\text{ so }\sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}}\\ v_{min}=\frac{8}{2\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57{\mathrm{ms}}^{-1}\end{array}\end{array}$