Question

A $2m$ wide truck is moving with a uniform speed $v_0=8m{s}^{-1}$ along a straight horizontal road. A pedestrian 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. $2.62m{s}^{-1}$
• B. $4.6m{s}^{-1}$
• C. $3.57m{s}^{-1}$
• D. $1.414m{s}^{-1}$

Answer 1

Answer: (C)

$3.57m{s}^{-1}$

Let the man start crossing the road at an angle $\theta$ with the road side. For safe crossing, the condition is that the man must cross the road by the time, the truck describes the distance $(4+2\cos \theta )$.
So, $\frac{4+2\cot \theta }{2}=\frac{2/\sin \theta }{v}$ or $v=\frac{2}{2\sin \theta +\cos \theta }$
For minimum $v,\frac{dv}{d\theta }=0$
$\Rightarrow \frac{-2(2\cos \theta -\sin \theta )}{(2\sin \theta +\cos \theta {)}^2}=0$
$\Rightarrow 2\cos \theta -\sin \theta =0$
$\Rightarrow \tan \theta =2,so\sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}}$
$v_{min}=\frac{2}{2\left(2/\sqrt{5}\right)+\left(1/\sqrt{5}\right)}=\frac{2\sqrt{5}}{5}=0.89m/s$.