Question

A $2m$ wide truck is moving with a uniform speed $v_0=8m/s$ along a straight horizontal road. A pedestrain starts to cross the road with a uniform speed $v$ when the truck is $4m$ away from him. The minimum value of $v$ so that he can cross the road safely is

• A. $2.62m/s$
• B. $4.6m/s$
• C. $3.57m/s$
• D. $1.414m/s$

Answer 1

The correct option is C $3.57m/s$

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 $v=\frac{8}{2\sin \theta +\cos \theta }.....\left(i\right)$
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 $\tan \theta =2$
For equation (i),
$v_{min}=\frac{8}{2\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57m/s$.