BITSAT 2015 Paper

Let the man starts crossing the road at an angle θ 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 + 2cot θ ∴ $\frac{4+2\cot\theta}{8}$ = $\frac{\frac{2}{\sin\theta}}{v}$ or v = $\frac{8}{2\sin\theta+\cos\theta}$ ... (i) 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 θ - sin t = 0 or tan θ = 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 m/s