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