# application of vectors: sailboat floats in a current

A sailboat floats in a current that flows due east at 1 meter per second. Due to a wind the boats actual speed relative to the shore is $({\sqrt 3})$ meters per second in a direction 30 degrees North of East. Find the speed and direction of the wind.

So far I have found the speed of the wind by using the formula for the resultant vector and got the speed to be 1 meter per second. Now how do I go about finding the direction of the wind? Can someone provide a step by step explanation. I don't understand why the wind would be East of North?

The Actual velocity $(\vec{v_r})$ rel to shore will be resultant of sailboat velocity $(\vec{v_b})$ and wind velocity $(\vec{v_w})$.

Figure 1:

Figure 2:

$\theta = 180 -30 = 150^\circ$

\begin{align} \vec{v_r} &= \vec{v_w}+\vec{v_b} \\ \\ \implies \vec{v_w} &= \vec{v_r}-\vec{v_b} \end{align}

\begin{align}|\vec{v_w}| &= \sqrt{|\vec{v_r}|^2 + |\vec{v_b}|^2 +2|\vec{v_r}||\vec{v_b}| \cos\theta} \\ \\ &=\sqrt{3 + 1 - 2\dfrac{3}{2}} \\ &=1 \end{align}

\begin{align} \tan\phi &= \dfrac{|\vec{v_b}|\sin\theta}{|\vec{v_r}|+|\vec{v_b}|\cos\theta} \\ \\ &= \dfrac{1/2}{\sqrt{3}-\frac{\sqrt{3}}{2}} \\ &= \frac{1}{\sqrt{3}}\\ \\ \implies \phi &= 30^\circ \end{align}

Therefore direction of wind is $30^\circ$ east of north, or $60^\circ$ north of east.

• after getting the value 30 how did you know it automatically had to be east of north though?
– Lil
Feb 1, 2016 at 18:23
• @Lil you can either use law of cosines, or the angle formula of $\tan\phi$. In that formula, $\phi$ is the angle made from first vector, first in the sense the one which comes first in the order of addition. So $\phi$ is angle made between $\vec{v_w}$ and $\vec{v_r}$ and not the angle made between $\vec{v_w}$ and $-\vec{v_b}$ Feb 2, 2016 at 4:34
• and $\vec{v_r}$ is $30^\circ$ north of east, and $\vec{v_w}$ is $30^\circ$ north of $\vec{v_r}$, So $\vec{v_w}$ is $60^\circ$ north of east, or $30^\circ$ east of north. Feb 2, 2016 at 4:39