21. I A kayaker needs to paddle north across a -wide harbor. The tide is going out, creating a tidal current that flows to the east at$2.0\text{m}/\text{s}$ . The kayaker can paddle with a speed of $3.0\text{m}/\text{s}$.

a. In which direction should he paddle in order to travel straight across the harbor?

b. How long will it take him to cross?

The kayaker needs to paddle across a $100\text{m}$wide harbor, the speed of the tidal current flowing in the east direction is $2.0\text{m}/\text{s}$, the speed of kayaker is $3.0\text{m}/\text{s}$.

Because the river is flowing east and the kayaker needs to paddle across, the kayaker should paddle in a northwest direction, at an angle of $\theta$to the north

$\theta ={\sin }^{-1}\left(\frac{v_r}{v_b}\right)$

Here, $v_r$is the speed of the tidal current $=2\text{m}/\text{s}$, and $v_b$ is the speed of the boat$=3\text{m}/\text{s}$

$\theta ={\sin }^{-1}\left(\frac{2\text{m}/\text{s}}{3\text{m}/\text{s}}\right)={41.8}^{°}$

Thus, the direction of kayaker is${41.81}^{°}$ towards the North-West direction.

Part (b)

The time it takes to cross the harbor is$t=\frac{\text{distance}}{\text{relativespeed}}$

The relative speed between the tidal current and the boat is

$\begin{array}{l}{v}_{\text{rel}}=\sqrt{v_b^2-v_r^2}=\sqrt{{(3\text{m}/\text{s})}^2-{(2\text{m}/\text{s})}^2}\\ =\sqrt{\frac{5{\text{m}}^2}{{\text{s}}^2}}\\ =2.236\text{m}/\text{s}\end{array}$

The time taken by kayaker to cross the harbour is

$t=\frac{\text{distasnce}}{v_{\text{rel}}}=\frac{100\text{m}}{2.236\text{m}/\text{s}}=44.7\text{s}$

Thus, the time taken by kayaker to cross the harbour is$44.7\text{s}$