Two cities lie along a river, d = 1 km apart. A boat moves…

Two cities lie along a river, $d = 1\ \mathrm{km}$ apart. A boat moves with speed $v_1 = 8\ \mathrm{km/h}$ relative to the water. By opening and closing a dam, the water in one case flows with speed $v_2 = 2\ \mathrm{km/h}$, and in the other case it is still. Compare the times the boat needs to travel the route upstream and back downstream in both cases.

Step 1: Flowing water

Upstream the current slows the boat down; downstream it speeds it up: \[t_{\text{up}} = \frac{d}{v_1 - v_2} = \frac{1\ \mathrm{km}}{6\ \mathrm{km/h}} = \frac{1}{6}\ \mathrm{h} = 10\ \mathrm{min},\]

\[ t_{\text{down}} = \frac{d}{v_1 + v_2} = \frac{1\ \mathrm{km}}{10\ \mathrm{km/h}} = 6\ \mathrm{min}. \]

\[ t = \frac{1}{6} + \frac{1}{10} = \frac{4}{15}\ \mathrm{h} = 16\ \mathrm{min}. \]

Step 2: Still water

\[ t_{\text{up}}' = t_{\text{down}}' = \frac{d}{v_1} = \frac{1}{8}\ \mathrm{h} = 7.5\ \mathrm{min}, \]

\[ t' = \frac{2d}{v_1} = \frac{1}{4}\ \mathrm{h} = 15\ \mathrm{min}. \]

Step 3: Ratios of the times

Upstream vs downstream: with the current $\dfrac{t_{\text{up}}}{t_{\text{down}}} = \dfrac{10}{6} = \dfrac{5}{3}$; in still water the two times are equal (ratio 1).

Round trip with the current vs in still water: $\frac{t}{t'} = \frac{4/15}{1/4} = \frac{16}{15}$.

\[\begin{gathered} \frac{t}{t'} = \frac{16}{15} \approx 1.07 \\ \frac{t_{\text{up}}}{t_{\text{down}}} = \frac{5}{3} \ \text{(flowing)}, \quad 1 \ \text{(still)} \end{gathered}\]

General formula

\[ \frac{t}{t'} = \frac{\dfrac{d}{v_1 - v_2} + \dfrac{d}{v_1 + v_2}}{\dfrac{2d}{v_1}} = \frac{v_1^2}{v_1^2 - v_2^2} = \frac{64}{60} = \frac{16}{15}\ \checkmark \]