A motorboat crosses a river of width d = 1 km . Its speed…

A motorboat crosses a river of width $d = 1\ \mathrm{km}$. Its speed relative to the water is $v_1 = 4\ \mathrm{km/h}$ and the river flows at $v_2 = 2\ \mathrm{km/h}$. Find: (a) the angle of the boat's path if it is pointed perpendicular to the current; (b) the direction in which the boat must be pointed so that its path is perpendicular to the current; (c) the ratio of the crossing times in the two cases.

Explanation of Method

Crossing a river: two ways to point the boat.

Step 1: (a) Pointing straight across

The boat's own velocity $v_1$ is straight across and the current adds $v_2$ along the river. The two are perpendicular, so the resulting velocity is

\[ v = \sqrt{v_1^2 + v_2^2} = \sqrt{16 + 4} = \sqrt{20} \approx 4.47\ \mathrm{km/h}, \] and the path makes an angle $\theta$ with the straight-across direction: \[\tan\theta = \frac{v_2}{v_1} = \frac{2}{4} = 0.5 \;\Longrightarrow\; \theta \approx 26.6^\circ.\]

\[\begin{gathered} \theta \approx 26.6^\circ \text{ downstream from straight across} \\ (63.4^\circ \text{ to the bank}) \end{gathered}\]

The crossing takes $t_1 = d / v_1 = 0.25\ \mathrm{h}$, during which the boat drifts $v_2 t_1 = 0.5\ \mathrm{km}$ downstream.

$\text{(a) heading straight across}$ $\text{(b) path straight across}$ $v_1$ $v_2$ $\theta$ $0.5\ \mathrm{km}$ $t_1 = 15\ \mathrm{min}$ $v_1$ $v_2$ $\alpha$ $t_2 \approx 17.3\ \mathrm{min}$ $\text{current}$ $d = 1\ \mathrm{km}$

boat relative to water
current
path (a)
path (b)

Drawn to scale. (a) The boat points straight across and is carried 0.5 km downstream. (b) It points 30° upstream so that the current's push is cancelled and it lands exactly opposite.

Step 2: (b) Path straight across

Point the boat at an angle $\alpha$ upstream. The upstream part of its own velocity must cancel the current: \[v_1 \sin\alpha = v_2 \;\Longrightarrow\; \sin\alpha = \frac{2}{4} = 0.5 \;\Longrightarrow\; \alpha = 30^\circ.\]

\[\begin{gathered} \alpha = 30^\circ \text{ upstream from straight across} \\ (60^\circ \text{ to the bank, upstream}) \end{gathered}\]

What is left of the boat's velocity for crossing is

\[ v_\perp = v_1 \cos\alpha = 4 \cdot \frac{\sqrt3}{2} \approx 3.46\ \mathrm{km/h}. \]

Step 3: (c) Ratio of crossing times

\[ t_1 = \frac{d}{v_1} = \frac{1}{4}\ \mathrm{h} = 15\ \mathrm{min}, \]

\[ t_2 = \frac{d}{v_1\cos\alpha} = \frac{1}{3.46}\ \mathrm{h} \approx 17.3\ \mathrm{min}. \]

\[ \frac{t_2}{t_1} = \frac{1}{\cos\alpha} = \frac{2}{\sqrt 3}. \]

\[\frac{t_2}{t_1} = \frac{2}{\sqrt3} \approx 1.15\]

Result

\[\boxed{\,\begin{gathered} \theta \approx 26.6^\circ \text{ downstream from straight across} \\ (63.4^\circ \text{ to the bank}) \end{gathered}\,}\]

\[\boxed{\,\begin{gathered} \alpha = 30^\circ \text{ upstream from straight across} \\ (60^\circ \text{ to the bank, upstream}) \end{gathered}\,}\]

\[\boxed{\,\frac{t_2}{t_1} = \frac{2}{\sqrt3} \approx 1.15\,}\]

Remark

Note. Pointing straight across is always the fastest way to cross, because all of the boat's own speed goes into crossing; the price is landing downstream. Landing exactly opposite costs time, and it is only possible when the boat is faster than the current ($v_1 \gt v_2$).