Answer:
[tex]41.81^{\circ}[/tex]
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 [tex]\overrightarrow{OA}[/tex] 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, [tex]\vec {R}[/tex] must be in the north direction.
Let [tex]\overrightarrow{AB}[/tex] 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 [tex]\overrightarrow {OA}[/tex] and head of the vector [tex]\overrightarrow{AB}[/tex] must lie on the north-south line.
Now, for this condition, from the triangle OAB
[tex]|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|[/tex]
[tex]\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3[/tex]
[tex]\Rightarrow \theta=\sin^{-1}\frac23[/tex]
[tex]\Rightarrow \theta=41.81^{\circ}[/tex]
Hence, the kayaker must paddle in the direction of [tex]41.81^{\circ}[/tex] in the north of east direction.
