An airplane flies with an airspeed of 50.0 m/s [E 40° N]. If its velocity relative to the ground is 30.0 m/s [SE], what is the wind velocity?

What you are being asked

You are given the airplane's velocity relative to the air (airspeed) and its velocity relative to the ground. The difference between those vectors is the wind velocity (air relative to ground).

Write the vector relationship

Let east be $+x$ and north be $+y$.

The velocity relationship is: $$ \vec v_{g}=\vec v_{a}+\vec v_{w} $$ So, $$ \vec v_{w}=\vec v_{g}-\vec v_{a} $$

Convert each given velocity into components

Airspeed: $50.0\,\text{m/s}\,[E\,40^\circ\, N]$

Angle is $40^\circ$ north of east: $$ \vec v_{a}=\langle 50\cos 40^\circ,\;50\sin 40^\circ\rangle =\langle 38.30,\;32.14\rangle\,\text{m/s} $$

Ground velocity: $30.0\,\text{m/s}\,[SE]$

$[SE]$ means $45^\circ$ south of east: $$ \vec v_{g}=\langle 30\cos 45^\circ,\;-30\sin 45^\circ\rangle =\langle 21.21,\;-21.21\rangle\,\text{m/s} $$

Subtract to get the wind vector

$$ \vec v_{w}=\vec v_{g}-\vec v_{a} =\langle 21.21-38.30,\;-21.21-32.14\rangle =\langle -17.09,\;-53.35\rangle\,\text{m/s} $$ The negative signs mean the wind blows toward the west and toward the south.

Convert wind components to speed and direction

Magnitude: $$ |\vec v_{w}|=\sqrt{(-17.09)^2+(-53.35)^2}\approx 56.0\,\text{m/s} $$

Direction (measured south of west): $$ \theta=\tan^{-1}\left(\frac{53.35}{17.09}\right)\approx 72.2^\circ $$ So the wind is $56.0\,\text{m/s}\,[W\,72^\circ\, S]$, which is the same as $56.0\,\text{m/s}\,[S\,18^\circ\, W]$.

Quick check

Adding $\langle -17.09,-53.35\rangle$ to the airspeed $\langle 38.30,32.14\rangle$ gives $\langle 21.21,-21.21\rangle$, which matches the stated $30.0\,\text{m/s}\,[SE]$.