Relative velocity in two dimensions

Slanted rain

You sit by the window of a moving bus. Outside, the rain falls straight down. Yet the drops run across your window in slanting lines. Who is right, you or the person standing at the bus stop?

Both are right. Each of you sees the rain relative to yourself. In this lesson we learn to switch between those points of view.

Relative velocity

The velocity of A relative to B is how A seems to move to someone who moves with B:

You met this along a line in the last chapter. In a plane it is the same rule, but now we subtract vectors. Also, $\vec{v}_{BA} = -\vec{v}_{AB}$: B sees A move one way, and A sees B move exactly the opposite way.

Worked example. A cyclist rides east at $12\ \text{m/s}$, and a scooter goes north at $5\ \text{m/s}$. How does the cyclist move, as seen by the scooter rider?

1. $\vec{v}_{\text{cyclist, scooter}} = \vec{v}_{\text{cyclist}} - \vec{v}_{\text{scooter}} = 12\hat{i} - 5\hat{j}\ \text{m/s}$.
2. Size: $\sqrt{(12\ \text{m/s})^2 + (5\ \text{m/s})^2} = 13\ \text{m/s}$.
3. Direction: $\tan^{-1}\dfrac{5\ \text{m/s}}{12\ \text{m/s}} = 22.6^\circ$ south of east. To the scooter rider, who moves north, the cyclist seems to drift backwards to the south as well.

Rain and an umbrella

Rain falls straight down at $8\ \text{m/s}$. You run at $6\ \text{m/s}$. The rain relative to you is $\vec{v}_{\text{rain}} - \vec{v}_{\text{you}}$. It has a downward part of $8\ \text{m/s}$ and a part of $6\ \text{m/s}$ that comes towards your face.

So, to you, the rain arrives at $\sqrt{(8\ \text{m/s})^2 + (6\ \text{m/s})^2} = 10\ \text{m/s}$, slanting at $\tan^{-1}\dfrac{6\ \text{m/s}}{8\ \text{m/s}} = 36.9^\circ$ from the vertical. Tilt the umbrella forward by that angle. The faster you run, the more you must tilt it.

A boat in a river

A boat moves at $v_b$ through the water. The water itself moves at $v_r$. The velocity of the boat over the ground is the sum:

Steer the boat yourself. The red arrow is the boat's own velocity in the water, the blue arrow is the river, and the amber arrow is the sum: where the boat really goes.

There are two classic questions. Take a river $d$ wide.

1. The quickest crossing. Only the part of the velocity across the river carries the boat to the other bank. It is largest when the boat points straight across. Then

The river still carries the boat downstream during that time. But it does not slow the crossing. The two directions are independent, as in the last lesson.

2. Landing straight opposite. Now the boat must point upstream at an angle $\phi$, so that its upstream part cancels the river: $v_b\sin\phi = v_r$. The part left for crossing is $v_b\cos\phi = \sqrt{v_b^2 - v_r^2}$, so

This crossing always takes longer than the quickest one.

Worked example. A river is $150\ \text{m}$ wide and flows at $1.5\ \text{m/s}$. A boat moves at $2.5\ \text{m/s}$ in still water.

1. Quickest: $t = \dfrac{150\ \text{m}}{2.5\ \text{m/s}} = 60\ \text{s}$, and it lands $1.5\ \text{m/s} \times 60\ \text{s} = 90\ \text{m}$ downstream.
2. Straight opposite: $\sin\phi = \dfrac{1.5\ \text{m/s}}{2.5\ \text{m/s}} = 0.6$, so $\phi = 36.9^\circ$ upstream. The crossing part is $\sqrt{(2.5\ \text{m/s})^2 - (1.5\ \text{m/s})^2} = 2\ \text{m/s}$, so $t = \dfrac{150}{2} = 75\ \text{s}$.

An edge case. What if the river is faster than the boat, $v_r > v_b$? Then $\sin\phi = v_r/v_b$ is more than $1$, which is impossible. The boat cannot land straight opposite. The best it can do is to keep the drift as small as possible. That happens when it points upstream at an angle $\alpha$ to the bank with $\cos\alpha = \dfrac{v_b}{v_r}$. Try it in the simulation with $v_r = 5\ \text{m/s}$ and $v_b = 3\ \text{m/s}$.

An aeroplane in the wind

A plane works like the boat, with the wind in place of the river. A pilot wants to fly due north at $360\ \text{km/h}$ through the air, while a wind blows towards the east at $90\ \text{km/h}$.

She must point the plane partly west, so that its westward part cancels the wind: $360\ \text{km/h}\sin\theta = 90\ \text{km/h}$, so $\sin\theta = 0.25$ and $\theta = 14.5^\circ$ west of north. Her speed over the ground is then $360\ \text{km/h}\cos 14.5^\circ = 349\ \text{km/h}$, a little less than her speed through the air.

Practise with new numbers

Try it in Python

The program tries several headings for the boat of the worked example, and prints the crossing time and the drift for each.

Remember these

• $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$, and $\vec{v}_{BA} = -\vec{v}_{AB}$.
• Boat over the ground = boat in the water + water. Plane over the ground = plane in the air + wind.
• Quickest crossing: point straight across, $t = d/v_b$. Straight opposite: point upstream with $\sin\phi = v_r/v_b$.
• To keep dry, tilt the umbrella towards the direction the rain seems to come from.

Check yourself

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

1. Two trains run on parallel tracks at $20\ \text{m/s}$ and $15\ \text{m/s}$. What is the speed of one relative to the other, if they go the same way? If they go opposite ways?

2. Rain falls straight down at $5\ \text{m/s}$. A girl runs at $5\ \text{m/s}$. At what angle should she tilt her umbrella?

3. A boat points straight across a river. On a day when the river flows faster, does the crossing take longer?

4. A car drives north at $10\ \text{m/s}$. A truck drives east at $10\ \text{m/s}$. How does the truck move, as seen by the car's driver?