Question

A ship is moving towards N-E with a velocity of 10 km/hr and to a passenger on board the wind appears to blow from north with a velocity of 10 root 2 km/hr .Actual velocity of the wind is _____.the answer is 10 km/hr towards S-E.explain how?​

Answer 1

Answer:

The actual velocity of the wind is 10 km/h in the S-E direction

Explanation:

If we take the east direction as positive x-axis and the north direction as positive y-axis then the velocity of the ship can be written as

$\vec v_S=10\cos45^\circ \hat i+10\sin45^\circ \hat j$

$\implies \vec v_S=\frac{10}{\sqrt{2}} \hat i+\frac{10}{\sqrt{2}} \hat j$

$\implies \vec v_S=5\sqrt{2} \hat i+5\sqrt{2}\hat j$

Since passenger is on board the ship

Therefore, velocity of the ship = velocity of the passenger

The wind appears to blow from the north to the south

i.e. velocity of the wind w.r.t. ship is is in -y direction and its magnitude is 10√2

Therefore,

$\vec v_{W/S}=-10\sqrt{2}\hat j$

From relative velocity we know that

$\boxed{v_{W/S}=v_{W/E}+v_{E/S}}$        (W/E means Wind relative to Earth)

$\implies v_{W/S}=v_{W/E}-v_{S/E}$

$\implies v_{W/S}=v_{W}-v_{S}$

$\implies v_{W}=v_{W/S}+v_{S}$

$\implies v_{W}=-10\sqrt{2}\hat j+5\sqrt{2} \hat i+5\sqrt{2}\hat j$

$\implies v_{W}=5\sqrt{2} \hat i-5\sqrt{2}\hat j$

Thus $+\hat i$ and $-\hat j$ component of the velocity indicates South-East direction

Magnitude of the velocity

$=\sqrt{(5\sqrt{2})^2+(-5\sqrt{2})^2}$

$=\sqrt{50+50}$

$=\sqrt{100}$

$=10$ km/h

Hope this helps.