Question

A swimmer wants to cross a river from point A to point B. Line AB makes an angle of 30° with the flow of the river. The magnitude of the velocity of the swimmer is the same as that of the river. The angle $\theta$ with the line AB should be ____°, so that the swimmer reaches point B.

Answer 1

30

Answer 2

Let $v_s$ be the velocity of the swimmer relative to the water, and $v_r$ be the velocity of the river flow. We are given that $|v_s| = |v_r| = v$. Let $v_{res}$ be the resultant velocity of the swimmer relative to the ground. The swimmer reaches point B, so $v_{res}$ must be directed along the line AB.

Let the direction of the river flow be along the positive x-axis. So, $v_r = v \hat{i}$. The line AB makes an angle of 30° with the flow of the river, so the direction of $v_{res}$ is at 30° with the positive x-axis. Let the swimmer's velocity relative to water, $v_s$, make an angle $\phi$ with the positive x-axis. Thus, $v_s = v \cos\phi \hat{i} + v \sin\phi \hat{j}$.

The resultant velocity is $v_{res} = v_s + v_r$. $v_{res} = (v \cos\phi \hat{i} + v \sin\phi \hat{j}) + v \hat{i} = (v \cos\phi + v) \hat{i} + v \sin\phi \hat{j}$.

Since $v_{res}$ is in the direction of 30° with the x-axis, the ratio of its y-component to its x-component must be equal to $\tan(30^\circ)$: $\frac{v_{res, y}}{v_{res, x}} = \tan(30^\circ)$ $\frac{v \sin\phi}{v \cos\phi + v} = \frac{1}{\sqrt{3}}$ Cancel $v$ from the numerator and denominator: $\frac{\sin\phi}{\cos\phi + 1} = \frac{1}{\sqrt{3}}$ Using the half-angle identities, $\sin\phi = 2 \sin(\phi/2)\cos(\phi/2)$ and $\cos\phi + 1 = 2 \cos^2(\phi/2)$: $\frac{2 \sin(\phi/2)\cos(\phi/2)}{2 \cos^2(\phi/2)} = \tan(\phi/2)$ So, $\tan(\phi/2) = \frac{1}{\sqrt{3}}$ This gives $\phi/2 = 30^\circ$, which implies $\phi = 60^\circ$. Thus, the swimmer's velocity relative to water ($v_s$) makes an angle of 60° with the river flow.

The angle $\theta$ is given as the angle between the line AB and the swimmer's velocity $v_s$. The line AB makes an angle of 30° with the river flow. The swimmer's velocity $v_s$ makes an angle of 60° with the river flow. From the diagram, $\theta$ is the difference between these two angles: $\theta = |\phi - 30^\circ| = |60^\circ - 30^\circ| = 30^\circ$