Let the man starts crossing the road at an angle \(\theta\) as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance \(4+A C\) or \(4+2 \cot \theta\).

\(\therefore \frac{4+2 \cot \theta}{8}=\frac{2 / \sin \theta}{v}\)

or \(\quad v=\frac{8}{2 \sin \theta+\cos \theta}\)

For minimum \(v, \frac{d v}{d \theta}=0\)

or \(\frac{-8(2 \cos \theta-\sin \theta)}{(2 \sin \theta+\cos \theta)^2}=0\)

or \(2 \cos \theta-\sin \theta=0\)

or \(\tan \theta=2\)

From equation (i),

\(v_{\min }=\frac{8}{2\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57\,m / s\)