Relative motion: steering a boat across a river

Crossing a river

Which way should a boat point to cross fastest, and which way to land straight across?

Set the boat's speed, its heading and the current. The boat's velocity over the ground is its velocity through the water plus the water's: watch the triangle of arrows decide the crossing time and how far downstream it lands.

High school
Adding velocities
Shortest time
Shortest path
Frames of reference

v_{\text{boat}}

3.0

\theta\,(^\circ)

90°

v_{\text{water}}

2.0

t

12.00

\vec v_{\text{ground}} = \vec v_{\text{boat}} + \vec v_{\text{water}}

$t_{\text{cross}}$: 2.00
$\text{drift}$: 4.00
$v_{\text{ground}}$: 3.61

Landingoff to one side

Guided tour

1Point straight across
The river is $6\,\text{m}$ wide. Rowing at $3\,\text{m/s}$ straight across takes $6/3=2\,\text{s}$ , but the $2\,\text{m/s}$ current carries the boat $2\times2=4\,\text{m}$ downstream in that time.
2A faster current
Double the current to $4\,\text{m/s}$ . The crossing still takes $2\,\text{s}$ , because the current only pushes sideways; the boat just lands $8\,\text{m}$ downstream. Pointing straight across is always the fastest crossing.
3Aim upstream to land opposite
Back to a $2\,\text{m/s}$ current, and turn the bow to $132^\circ$ , angled upstream. The upstream part of the boat's velocity, $3\cos48^\circ\approx2\,\text{m/s}$ , cancels the current and the boat lands almost exactly opposite. The price: the crossing now takes about $2.69\,\text{s}$ .
4When the river wins
Now the boat makes $2\,\text{m/s}$ in a $3\,\text{m/s}$ current. No heading can cancel the current, so the boat always lands downstream. At $132^\circ$ the drift is about $6.71\,\text{m}$ , the smallest it can be.

Explanation

1
Velocities add as arrows
The boat moves through the water; the water moves over the ground. Its velocity relative to the ground is the sum of the two, added tip to tail:
$\vec v_{\text{ground}}=\vec v_{\text{boat}}+\vec v_{\text{water}}$
On the board the green arrow (boat through water) and the blue arrow (water) form a triangle with the amber one, the path the boat really follows.
2
The crossing time
Only the part of the velocity across the river gets the boat to the other side. With the bow at angle $\theta$ to the bank:
$t=\frac{d}{v_{\text{boat}}\sin\theta}$
The current does not appear: it is parallel to the bank. The time is shortest at $\theta=90^\circ$ , $t_{\min}=\dfrac{d}{v_{\text{boat}}}$ .
3
The shortest path
To land straight across, the ground velocity must point straight across, so the boat's upstream component must equal the current: $v_{\text{boat}}\cos\alpha=v_{\text{water}}$ , where $\alpha$ is the angle to the upstream bank. That is only possible when the boat is faster than the water. If it is slower, the smallest drift comes from $\cos\alpha=\dfrac{v_{\text{boat}}}{v_{\text{water}}}$ , with the ground velocity at right angles to the boat's.

Check yourself

Q01

A river is $100\,\text{m}$ wide. A boat makes $5\,\text{m/s}$ in still water; the current is $3\,\text{m/s}$ . What is the shortest crossing time?

Q02

Same river and boat. To land straight across, how long does the crossing take?

Q03

If the current suddenly gets faster mid-river while the bow stays pointed straight across, the crossing time

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