A $2 \mathrm{~m}$ wide truck is moving with a uniform speed $v_{0}=8 \mathrm{~m} / \mathrm{s}$ along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed $v$ when the truck is $4 \mathrm{~m}$ away from him. The minimum value of $v$ so that he can cross the road safely is

Question

Aicharya · Accepted Answer

For safe crossing, the condition is that the man must cross the road by the time the truck covers 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 $v, \frac{d v}{d \theta}=0 \quad \Rightarrow \quad \tan \theta=2$
From equation (i), $\quad v_{\min }=\frac{8}{\sqrt{5}}=3.57 \mathrm{~m} / \mathrm{s}$ *

A $2 \mathrm{~m}$ wide truck is moving with a uniform speed $v_{0}=8 \mathrm{~m} / \mathrm{s}$ along a…

Solution