In a village game, three boys A, B and C are in motion. The velocity of boy A seen by boy B is $1 m s^{- 1}$ towards the north – east direction. The velocity of the boy B as seen by boy C is $1 m s^{- 1}$ towards the North – West direction. The direction of velocity of boy C seen by boy A is

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$1 m s^{- 1}$ towards East

$\sqrt{2} m s^{- 1}$ towards North

$1 m s^{- 1}$ towards South

$\sqrt{2} m s^{- 1}$ towards South

answer is C.

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Detailed Solution

Here, $v_{m} = 10 \hat{i}$
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and $v_{b} = 5 \sin 30^{0} \hat{i} + 5 \cos 30^{0} \hat{j} v_{b m} = v_{b} - v_{m}$
$= 5 \sin 30^{0} \hat{i} + 5 \cos 30^{0} \hat{j} = - 7.5 \hat{i} + 2.5 \sqrt{3} \hat{j}$

$\tan θ = \frac{7.5}{2.5 \sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} θ = 60^{0}$
$| v_{b m} | = \sqrt{{(- 7.5)}^{2} + {(2.5 \sqrt{3})}^{2}} = \sqrt{75} = 5 \sqrt{3} k m h^{- 1}$