Question

A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the east at 2.0 m/s. The kayaker can paddle with a speed of 3.0 m/s. In which direction should he paddle in order to travel straight across the harbor? Express your answer in degrees measured north of east.

Answer 1

Answer:

$41.81^{\circ}$

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector $\overrightarrow{OA}$ is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, $\vec {R}$ must be in the north direction.

Let $\overrightarrow{AB}$ is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector $\overrightarrow {OA}$ and head of the vector $\overrightarrow{AB}$ must lie on the north-south line.

Now, for this condition, from the triangle OAB

$|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|$

$\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3$

$\Rightarrow \theta=\sin^{-1}\frac23$

$\Rightarrow \theta=41.81^{\circ}$

Hence, the kayaker must paddle in the direction of $41.81^{\circ}$  in the north of east direction.