Question

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

 

Hint:

In relative velocity problems, drawing a vector triangle ($\vec{v}_{resultant} = \vec{v}_{relative} + \vec{v}_{frame}$) is often the clearest way to solve. Applying the Law of Sines or Cosines to this triangle can simplify the calculations.

Answer 1

Correct Answer: 30

Let $\vec{v}_{s,r}$ be the velocity of the swimmer with respect to the river.
Let $\vec{v}_{r,g}$ be the velocity of the river with respect to the ground.
Let $\vec{v}_{s,g}$ be the velocity of the swimmer with respect to the ground.
The relationship is $\vec{v}_{s,g} = \vec{v}_{s,r} + \vec{v}_{r,g}$.
We are given that the magnitude of the swimmer's velocity is the same as the river's velocity. This means $|\vec{v}_{s,r}| = |\vec{v}_{r,g}|$. Let this magnitude be $u$.
The river flows horizontally. Let's set this as the x-axis, so $\vec{v}_{r,g} = u\hat{i}$.
The swimmer wants to reach point B, so his resultant velocity $\vec{v}_{s,g}$ must be along the line AB, which makes an angle of 30$^\circ$ with the river flow.
Let the swimmer's heading (direction of $\vec{v}_{s,r}$) be at an angle $\beta$ with the river flow (x-axis).
$\vec{v}_{s,r} = u\cos\beta \hat{i} + u\sin\beta \hat{j}$.
The resultant velocity is:
$\vec{v}_{s,g} = (u\cos\beta + u)\hat{i} + (u\sin\beta)\hat{j}$.
The direction of this resultant velocity is 30$^\circ$. So, the tangent of the angle is:
$\tan(30^\circ) = \frac{v_{s,g,y}}{v_{s,g,x}} = \frac{u\sin\beta}{u(\cos\beta + 1)} = \frac{\sin\beta}{1 + \cos\beta}$.
Using trigonometric half-angle identities: $\sin\beta = 2\sin(\beta/2)\cos(\beta/2)$ and $1+\cos\beta = 2\cos^2(\beta/2)$.
$\tan(30^\circ) = \frac{2\sin(\beta/2)\cos(\beta/2)}{2\cos^2(\beta/2)} = \tan(\beta/2)$.
Therefore, $\beta/2 = 30^\circ \implies \beta = 60^\circ$.
This means the swimmer must head at an angle of 60$^\circ$ with respect to the river bank to achieve a resultant path at 30$^\circ$.
The question asks for the angle $\theta$ with the line AB.
The line AB is the direction of the resultant velocity $\vec{v}_{s,g}$ (at 30$^\circ$).
The swimmer's heading is the direction of $\vec{v}_{s,r}$ (at 60$^\circ$).
The angle $\theta$ between these two directions is:
$\theta = \beta - 30^\circ = 60^\circ - 30^\circ = 30^\circ$.