A 2-m wide truck is moving with a uniform speed v₀ = 8 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 v so that he can cross the road safely is

 Let the man start crossing the road at an angle 0 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 $\theta$) 

$$\begin{gathered} \frac{4+2\mathrm{cot\theta }}{8}=\frac{2/\mathrm{sin\theta }}{\mathrm{v}}\text{ or } \mathrm{v}=\frac{8}{2\mathrm{sin\theta }+\mathrm{cos\theta }} \\ \text{ For minimum }\mathrm{v},\frac{\mathrm{dv}}{\mathrm{d\theta }}=0 \\ \frac{-8(2\mathrm{cos\theta }-\mathrm{sin\theta })}{(2\mathrm{sin\theta }+\mathrm{cos\theta }{)}^2}=0 \text{ or } 2\mathrm{cos\theta }-\mathrm{sin\theta }=0 \\ \mathrm{tan\theta }=2,\text{ so }\mathrm{sin\theta }=\frac{2}{\sqrt{5}},\mathrm{cos\theta }=\frac{1}{\sqrt{5}} \\ {\mathrm{v}}_{\min }=\frac{8}{2\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57{ \mathrm{ms}}^{-1} \end{gathered}$$