A man walks for some time $^{'} t^{'}$ with velocity $(v)$ due east. Then he walks for same time $^{'} t^{'}$ with velocity $(v)$ due north. The average velocity of the man is

2 v

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\sqrt{2} v

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v

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\frac{v}{\sqrt{2}}

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Solution

The correct option is D $\frac{v}{\sqrt{2}}$
From the figure,
Displacement, $S = O B = \sqrt{(O A)^{2} + (A B)^{2}}$
$S = \sqrt{(v t)^{2} + (v t)^{2}}$
$S = \sqrt{2} v t m / s$
Total time, $T = 2 t$
Average velocity, $v_{a v g} = \frac{S}{T}$
$v_{a v g} = \frac{\sqrt{2} v t}{2 t}$
$v_{a v g} = \frac{v}{\sqrt{2}} m / s$
The correct option is D.

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