Question

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

• A. $\sqrt{5}$
• B. $\sqrt{\frac{5}{2}}$
• C. $\frac{8}{\sqrt{5}}$
• D. None

Answer 1

The correct option is C

$\frac{8}{\sqrt{5}}$

As the truck reaches from A to C, the man should reach C from B.
$\frac{2}{vcos\theta }=\frac{4+2tan\theta }{8}16=4vcos\theta +2sin\theta \Rightarrow v=\frac{1}{2cos\theta +sin\theta }Or{V}_{min}\frac{dv}{d\theta }=0\Rightarrow \frac{8(cos\theta -2sin\theta )}{(2cos\theta +sin\theta {)}^2}=0\Rightarrow tan\theta =\frac{1}{2}$

$V_{min}=\frac{8}{2\times \frac{2}{\sqrt{5}+\frac{1}{\sqrt{5}}}}=\frac{8}{\sqrt{5}}m{s}^{-1}$