Question

A river is flowing with a speed of $1{\text{kmh}}^{-1}$. A swimmer wants to go to point $\text{C}$ starting form $\text{A}$. He swims with a speed of $5{\text{kmh}}^{-1}$ at an angle $\theta$ with the river flow. If $\text{AB}=\text{BC}=400\text{m}$, at what angle with the river bank should the swimmer swim?

• A. $\theta =\frac{1}{2}{\sin }^{-1}\frac{24}{25}$
• B. $\theta =\frac{1}{2}{\sin }^{-1}\frac{12}{25}$
• C. $\theta =\frac{1}{2}{\sin }^{-1}\frac{16}{25}$
• D. $\theta =\frac{1}{2}{\sin }^{-1}\frac{9}{25}$

Answer 1

The correct option is A $\theta =\frac{1}{2}{\sin }^{-1}\frac{24}{25}$

As $\text{AB = BC}$, swimmer has to travel equal distances in both directions. This implies that the resultant path of swimmer is at ${45}^{\circ }$ with the river bank, therefore X- and Y- components of the swimmer's resultant must be equal.

Condition for reaching the point $\text{C}$ is
$\tan {45}^{\circ }=\frac{V_{mY}}{V_{mX}}$
$\Rightarrow V_{mX}=V_{mY}$


and we have, $V_{mX}=V_r+V_m\cos \theta$
$V_{mY}=V_m\sin \theta$

Therefore, to cover equal distances in both directions
$V_r+V_m\cos \theta =V_m\sin \theta$
$1+5\cos \theta =5\sin \theta$
$5\sin \theta -5\cos \theta =1$
Squaring both sides
$\Rightarrow ({\sin }^2\theta +{\cos }^2\theta )-\sin 2\theta =\frac{1}{25}$
$\Rightarrow \sin 2\theta =\frac{24}{25}$
or $\theta =\frac{1}{2}{\sin }^{-1}\frac{24}{25}$