A 2 m wide truck is moving with a uniform speed v_ 0 =8 m / s along...

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 $v=\frac{8}{2 \sin \theta+\cos \theta}$ 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 }=$\frac{8}{2\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57 \mathrm{ m}$/ $\mathrm{s}$