vectors
The water in a river boat moves south at 10 mi/hr. If a motorboat is traveling due east at a speed of 20 mi/hr relative to the shore determine the speed and direction of the boat relative to the moving water.
Can someone go through this problem step by step. Initially I constructed a right triangle where 10 mi/hr and 20 mi/hr represented the legs and found the hypotenuse but I don't think that is correct.
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.
Best Answer
Here is a picture:
The speed relative to water is parallel to the hypotenuse. So it is:
$$\sqrt{10^2+20^2}=10\sqrt{5}$$
The angle is:
$$\arctan{\frac{10}{20}}=\arctan{\frac{1}{2}}$$
to the north from the east.
So the driver has to point the boat to that direction with speed $10\sqrt{5}$ such that the resultant direction will be straight to the east.