Physics - Kinematics Question with Solution | TestHub

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 । $AC$ 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 $v,\frac{dv}{d\theta }=0$
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), $$v_{\min }=\frac{8}{2\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57\mathrm{m}/\mathrm{s}$$