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Question

In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ?

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Solution

Given: In a harbor, speed of wind is 72 kmh -1 , the flag on the mast of a boat anchored in the harbor flutters along the N-E direction and the boat starts moving at a speed of 51 kmh -1 to the north.

When the boat is anchored, the flag is fluttering in the north-east direction. Hence, wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity of the wind with respect to the boat.



The angle between wind speed and boat is given as,

θ= 90°+45° =135°

The angle between wind speed and relative velocity of the wind with respect to the boat is given as,

tanβ= v b sinθ v w + v b cosθ

Where, the speed of boat is v b and the speed of wind is v w .

By substituting the given values in the above expression,

tanβ= 51sin135 72+51cos135 = 51× 1 2 72+51×( 1 2 ) = 51 72 2 51 =1.0038

Further solving for β, we get

β= tan 1 ( 1.0038 ) =45.11°

Angle with respect to east direction is

45.11°45°=0.11°.

Thus, the flag will flutter almost due east.


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