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Q.

A 2m wide truck is moving with a uniform speed $v_{0} = 8 m s^{- 1}$ 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

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a

$2.62 m s^{- 1}$

b

$4.6 m s^{- 1}$

c

$3.57 m s^{- 1}$

d

$1.414 m s^{- 1}$

answer is C.

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Detailed Solution

Question Image

Let the man starts crossing the road at an angle $θ$ units the road side, for safe crossing the
condition is the the man must cross the road by the time the true describes the distance $4 + x$
$\frac{4 + x}{8} = \frac{\sqrt{4 + x^{2}}}{v} \Rightarrow v = \frac{8 \sqrt{4 + x^{2}}}{4 + x}$
For minima $\frac{d v}{d x} = 0 \Rightarrow x = 1 \Rightarrow v = \frac{8}{\sqrt{5}} m / s$
$= 3.57 m / s$

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