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A boat is moving towards east with velocity 4 m/s with respect to still water and river is flowing towards north with velocity 2 m/s and the wind is blowing towards north with velocity 6 m/s. The direction of the flag blown over by the wind hoisted on the boat is:

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north-west

south-east

$\tan^{- 1} (\frac{1}{2})$ with east

north

answer is A.

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Detailed Solution

${\vec{v}}_{B g} = {\vec{v}}_{B R} + {\vec{v}}_{R g} = 4 \hat{i} + 2 \hat{j} {\vec{v}}_{w g} = 6 \hat{j} {\vec{v}}_{w B} = {\vec{v}}_{w} - {\vec{v}}_{B} = 6 \hat{j} - 4 \hat{i} - 2 \hat{j} = - 4 \hat{i} + 4 \hat{j}$
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Direction will be north-west

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