Question

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?

Short answer

Part (a) The direction of kayaker is ${41.81}^{°}$towards the North-West direction.

Part (b) The time taken by kayaker to cross the harbour is$44.7\text{s}$

Step-by-step solution

Part (a)

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)

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$44.7\text{s}$