Relative Velocity in Two Dimensions

When the motion is not along one line

So far, every velocity lay along one line. Now let them point in different directions. The rule is the same, $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$, but now we add and subtract vectors: we put the arrows head to tail, and we use Pythagoras and trigonometry.

Rain and an umbrella

Rain falls straight down at $5\ \text{m/s}$. You walk at $2\ \text{m/s}$. Relative to you, the rain has its own $5\ \text{m/s}$ down, and it also moves $2\ \text{m/s}$ backward, because you move forward through it. So it seems to come at a slant, from in front of you.

• Its speed relative to you is $\sqrt{(5\ \text{m/s})^2 + (2\ \text{m/s})^2} = 5.39\ \text{m/s}$.
• Its angle from the vertical is $\theta$, with $\tan\theta = \dfrac{2\ \text{m/s}}{5\ \text{m/s}} = 0.4$, so $\theta = 21.8^\circ$.

So you tilt your umbrella forward by about $22^\circ$. The faster you walk, the more you must tilt it.

A boat crossing a river

A boat moves at $u$ through the water, and the river flows at $w$. Over the ground, the boat moves along the sum of the two velocities.

Heading straight across. Only the boat's own speed carries it over the river, so the time to cross a width $d$ is

Meanwhile the river carries it downstream by $w\,t$. This is the quickest crossing.

Landing right opposite. To land straight across, the boat must head upstream at an angle $\theta$, so that the upstream part of its velocity cancels the river: $u\sin\theta = w$. It now crosses with only $u\cos\theta$, so it takes longer: $t = \dfrac{d}{u\cos\theta}$. This needs $u > w$; a boat slower than the river cannot land opposite.

An example. A river is $100\ \text{m}$ wide and flows at $2\ \text{m/s}$. A boat moves at $4\ \text{m/s}$ in still water.

1. Straight across: $t = \dfrac{100\ \text{m}}{4\ \text{m/s}} = 25\ \text{s}$, and it lands $2\ \text{m/s} \times 25\ \text{s} = 50\ \text{m}$ downstream.
2. To land opposite: $\sin\theta = \dfrac{2\ \text{m/s}}{4\ \text{m/s}} = 0.5$, so $\theta = 30^\circ$ upstream. It crosses at $4\ \text{m/s} \times \cos 30^\circ = 3.46\ \text{m/s}$, which takes $\dfrac{100\ \text{m}}{3.46\ \text{m/s}} = 28.9\ \text{s}$.

You should find that the crossing time is shortest at a heading of $0^\circ$, and that the drift becomes zero when $u\sin\theta = w$.

Aircraft and wind

A pilot faces the same problem in a crosswind. To fly due north in a wind from the west, the aircraft must point a little to the west, into the wind, so that its westward part cancels the wind. Its speed over the ground is then less than its speed through the air.

Try it in Python

The program crosses the river of the example with four different headings.

Practise with new numbers

Remember these

• In two dimensions, relative velocities add and subtract as vectors.
• Rain falling straight down seems to slant towards a moving person, at $\tan\theta = \dfrac{v_{\text{person}}}{v_{\text{rain}}}$ from the vertical.
• A boat heading straight across crosses in the least time, $t = \dfrac{d}{u}$, and drifts $w\,t$ downstream.
• To land right opposite, head upstream with $u\sin\theta = w$; then $t = \dfrac{d}{u\cos\theta}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Rain falls straight down at $9\ \text{m/s}$. A cyclist rides at $12\ \text{m/s}$. How fast does the rain move relative to the cyclist?

2. A boat moves at $3\ \text{m/s}$ in still water and heads straight across a river $90\ \text{m}$ wide that flows at $1\ \text{m/s}$. How long does it take, and how far downstream does it land?

3. A swimmer is slower than the river. Can she land right opposite her starting point? Why?

4. Why does a pilot who wants to fly due north in a west wind point the aircraft a little to the west?