A ship moves along the equator to the east with a speed 30 km-h- Southeastern wind blows 60 - to the east with 15 km h -1- Find the wind velocity relative to the ship-

The correct option is A 41 km h ^ 1 ,tan ^ 1 √(3) 5 North of WestRelative velocity is v_w v_s ( 15cos60î + 15sin60 #160; ĵ) 30î so ...

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Byju's Answer

Standard XII

Physics

Relative Velocity

A ship moves ...

Question

A ship moves along the equator to the east with a speed $30 k m / h$ . Southeastern wind blows $60^{\circ}$ to the east with $15 k m h^{- 1}$ . Find the wind velocity relative to the ship.

41 k m h^{- 1}, {tan}^{- 1} \frac{\sqrt{3}}{5}

North of West

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23.7 k m h^{- 1}, {tan}^{- 1} \frac{1}{3}

North of West

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37.5 k m h^{- 1}, {tan}^{- 1} \frac{1}{5}

North of West

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none of these

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Solution

The correct option is A $41 k m h^{- 1}, {tan}^{- 1} \frac{\sqrt{3}}{5}$ North of West
Relative velocity is $v_{w} - v_{s}$
$(- 15 c o s 60^i + 15 s i n 60^j) - 30^i$

so $| V | = \sqrt{{37.5}^{2} + {12.99}^{2}} = 39.98 m / s e c$

and we know that $t a n θ = \frac{y}{x}$
$t a n θ = \frac{12.99}{37.5} = \frac{\sqrt{3}}{5}$ in North West

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A ship moves along the equator to the east with a speed 30 km/h. Southeastern wind blows 60∘ to the east with 15 kmh−1. Find the wind velocity relative to the ship.

The correct option is A 41 kmh−1,tan−1√35 North of WestRelative velocity is vw−vs(−15cos60^i+15sin60^j)−30^iso |V|=√37.52+12.992=39.98m/secand we know that tanθ=yxtanθ=12.9937.5=√35 in North West

A ship moves along the equator to the east with a speed $30 k m / h$ . Southeastern wind blows $60^{\circ}$ to the east with $15 k m h^{- 1}$ . Find the wind velocity relative to the ship.