A classical mechanics problem by Tushar Saxena

To have a minimum constant velocity $v$ , the pedestrian has to run across the road diagonally clearing 2 m (the width of the truck) perpendicular to the road and a distance $x$ along the direction of the truck is moving. Therefore, if he takes time $t$ to run the diagonal distance, then: $vt = \sqrt{x^2+2^2}$ And in time $t$ , the truck, with a constant speed of 8 m/s, would have clear a distance of $x+4$ . $8t = x+4$ Divide the two equations to get rid of $t$ , we get: $v = \frac {8\sqrt{x^2+4}}{x+4}$ Now, we differentiate $v(x)$ to get its minimum $v_{min}$ . $\frac {dv}{dx} = \frac{4(x^2+4)^{-\frac{1}{2}}(2x)}{x+4}-\frac{8(x^2+4)^{\frac{1}{2}}}{(x+4)^2}$ And $v=v_{min}$ when $\dfrac{dv}{dx}=0$ : $\Rightarrow \frac {8x}{(x+4)\sqrt{x^2+4}}=\frac{8\sqrt{x^2+4}}{(x+4)^2}$ $\Rightarrow x(x+4)=x^2+4$ $\Rightarrow 4x=4 \Rightarrow x=1$ Therefore, $v_{min} = \frac {8\sqrt{1+4}}{1+4}=\boxed {1.6\sqrt{5}}\quad m/s$

same as of chew seong