Under the conditions of the previous problem (Translational…

Under the conditions of the previous problem (Translational Kinematics 017): the cities are $d = 1\ \mathrm{km}$ apart, the boat's speed relative to the water is $v_1 = 8\ \mathrm{km/h}$, and the water either flows at $v_2 = 2\ \mathrm{km/h}$ or is still. Find the distance the boat travels relative to the water on the round trip in both cases.

Step 1: Distance relative to the water

Relative to the water, the boat always moves at $v_1$. So the distance it covers through the water is $v_1$ times the time it is moving. The times come from the previous problem.

Step 2: Flowing water

\[ d_{\text{up}} = v_1\,t_{\text{up}} = 8\ \mathrm{km/h} \cdot \frac{1}{6}\ \mathrm{h} = \frac{4}{3}\ \mathrm{km}, \]

\[ d_{\text{down}} = v_1\,t_{\text{down}} = 8 \cdot \frac{1}{10} = \frac{4}{5}\ \mathrm{km}. \]

\[ d_{\text{total}} = \frac{4}{3} + \frac{4}{5} = \frac{20 + 12}{15} = \frac{32}{15}\ \mathrm{km}. \]

Step 3: Still water

With no current, the water and the shore are the same reference, so the distance is just there and back: \[d' = v_1\,t' = 8\ \mathrm{km/h} \cdot \frac{1}{4}\ \mathrm{h} = 2\ \mathrm{km} = 2d.\]

\[\begin{gathered} d_{\text{total}} = \frac{32}{15}\ \mathrm{km} \approx 2.13\ \mathrm{km} \ \text{(flowing)} \\ d' = 2\ \mathrm{km} \ \text{(still)} \end{gathered}\]

Check another way

Upstream the water carries the boat back $ v_2 t_{\text{up}} = 2 \cdot \tfrac{1}{6} = \tfrac{1}{3}\ \mathrm{km} $, so through the water: $ 1 + \tfrac{1}{3} = \tfrac{4}{3}\ \mathrm{km}\ \checkmark $

Downstream it carries the boat forward $ v_2 t_{\text{down}} = 2 \cdot \tfrac{1}{10} = 0.2\ \mathrm{km} $, so through the water: $ 1 - 0.2 = 0.8\ \mathrm{km}\ \checkmark $