The operator of a boat wishes to cross a 5-kilometer-wide river that is flowing to the east at 10 meters per second.He wishes to reach the exact point on the opposite shore 15 minutes after starting.At what speed and in what direction should the boat travel?

$\begin{array}{1 1}\text{11.2m/s at 26.6}^{\large\circ}\text{ E of N}\\\text{8.66m/s at 63.4}^{\large\circ}\text{ W of N}\\\text{11.2m/s at 63.4}^{\large\circ}\text{ W of N}\\\text{8.66m/s at 26.6}^{\large\circ}\text{ E of N}\end{array}$

Answer : $\text{11.2m/s at 63.4}^{\large\circ}\text{ W of N}$
The river is flowing at 10m/s to the right(east) and the resultant desired velocity is 5m/s up (north).
Therefore,the actual velocity, relative to the river ,is heading W of N.
$\Rightarrow$ Using Pythagoras theorem, we can calculate the velocity of the boat relative to the water as $v_B = \sqrt { 10^2 + 5^2} = 11.2\;m/s$
The angle is given by the tangent function:
From the diagram above, $\tan\theta=\large\frac{10}{5}$$=2$.Therefore $\theta=63.4^{\large\circ}$ W of N.
edited Aug 13, 2014