Entrance Exams Q&A: Q. 1.8 -- "Problems in General Physics" by I.E. Irodov

$\textbf{Question}$ :

Two boats, and , move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines: the boat along the river, and the boat across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio of times of motion of boats $\tau_a/\tau_b$ if the velocity of each boat with respect to water is $\eta$ = 1.2 times greater than the stream velocity.

$\textbf{Answer}$ :

Assume that both boats cover a distance of in their excursion. Also, assume that the stream velocity is , then the velocity of each boat w.r.t water is $\eta v$ .

Time spent by boat = $\frac{d}{v(\eta+1)} + \frac{d}{v(\eta - 1)}$ = $\frac{2d}{v}\frac{\eta}{\eta^2 - 1}$

For boat to go across the river and come back, it must have traveled at an angle such that a component of its velocity counters the river flow.

Image diag

So the boat travels such that $\eta v\sin\phi = v$

$\begin{displaymath}\Rightarrow \sin\phi = \frac{1}{\eta} \end{displaymath}$

$\begin{displaymath}\Rightarrow \cos\phi = \sqrt{1-\frac{1}{\eta^2}} \end{displaymath}$

As the figure shows, the boat travels perpendicular to the river with a speed $\eta v\cos\phi$ . And thus the boat's speed perpendicular to stream is $\eta v\sqrt{1-\frac{1}{\eta^2}}$ = $v \sqrt{\eta^2 - 1}$ . Therefore, the time taken by boat to go a distance and come back is $\frac{2d}{v \sqrt{\eta^2 - 1}}$