Explanation

(c) Let the man starts 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 truck describes the distance(4+2 cot $\theta$),
So, $\frac{4 + 2cot\theta}{8} = \frac{2lsin\theta}{v}$
or $v = \frac{8}{2sin\theta = 2cos\theta}$
For minimum $v = \frac{dv}{d\theta} = 0$
or $\frac{- 8(2cos\theta - sin\theta)}{(2sin\theta + cos\theta)^{2}} = 0$
or $2cos\theta - sin\theta = 0$
or tan$\theta = 2,$ so, sin$\theta = \frac{2}{\sqrt{5}},$ $cos\theta$=$\frac{1}{\sqrt{5}}$
$\therefore\ v_{\min} = \frac{6}{2\left( \frac{2}{\sqrt{5}} \right) + \frac{1}{\sqrt{5}}}$
$= \frac{8}{\sqrt{5}} = 3.57ms^{- 1}$