Question

A river has a width d. A fisherman in a boat crosses the river twice. During the first crossing, his goal is to minimize the crossing time. During the second crossing, his goal is to minimize the distance that the boat is carried downstream. In the first case, the crossing time is $T_0$. In the second case, the crossing time is $3T_0$. What is the speed of the river flow? Find all possible answers.

Answer 1

Case I: If $v_R<v_B$, the boat can cross river along a path perpendicular to flow.
Case II: If $v_B<v_R$, drift cannot be zero, apply calculus in this case.
Case I: If $v_R<v_B$
Shortest path: $\frac{d}{v_{B}sin\theta }=3T_0$ ...(i)
Quickest path: $\frac{d}{v_B}=T_0$ ...(ii)
Also $v_R-v_{B}cos\theta =0$ for shortest path ...(iii)
Thus, $sin\theta =\frac{1}{3}$ from (i) and (ii)
or $v_R=v_{B}cos\theta$
$=\frac{d}{T_0}\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}d}{3t_0}$
Case II: If $v_B<v_R$
$x=\left(\frac{d}{v_{B}sin\theta }\right)(v_R-v_{B}cos\theta )$
$=\frac{d}{v_B}(v_{R}cosec\theta -v_{B}cot\theta )$
For minimum x, $\frac{dx}{d\theta }=-0$
$v_B(-cose{c}^2\theta )-v_{R}cosec\theta cot\theta =0$
$cos\theta =\frac{v_B}{v_R}$
Time taken in this case is given by $3T_0=\frac{V_{R}d}{V_B\sqrt{v_R^2-v_B^2}}$
Also $v_B$=$\frac{d}{T_0}$
On solving, we get $V_R=\sqrt{\frac{3}{2}}\frac{d}{T_0}$