Let the man starts crossing the road at an angle θ 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 + AC or 4 + 2 cot θ.


​$\therefore \frac{4+2\text{cot}\text{θ}}{8}=\frac{2/\text{sin}\text{θ}}{\text{v}}$

$\text{or}\text{v}=\frac{8}{\text{2}\text{sin}\text{θ}+\text{cos}\text{θ}}\text{...(1)}$

$\text{For minimum v,}\frac{\text{dv}}{\text{d}\text{θ}}=0$

$\text{or}\frac{-8\left(2\text{cos}\text{θ}-\text{sin}\text{θ}\right)}{(\text{2}\text{sin}\text{θ}+\text{cos}\text{θ}{)}^2}=0$

or                     2 cos θ - sin θ = 0

or                                   tan θ = 2



from Equation (1)

${\text{v}}_{\text{min}}=\frac{8}{\text{2}\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}} \mathrm{m} {\mathrm{s}}^{-1}$