Relative velocity along a line

What you see depends on how you move

You are in a car that goes at $60\ \text{km/h}$. Ahead of you, a truck also goes at $60\ \text{km/h}$. To you, the truck is not moving at all. It stays at the same distance in front of you.

Now another car passes you at $80\ \text{km/h}$. To you, it seems to go past slowly, at only $20\ \text{km/h}$.

The velocity of an object as seen from another moving object is called its relative velocity.

You have felt this many times. Two trains at a station, side by side, and one starts: you cannot tell which is moving. A train that comes towards you on the next track seems to rush past, much faster than a train that goes the same way as yours. A person who runs beside you at your speed seems to be standing still. In all of these, what you see is the relative velocity.

The rule

Choose one direction as positive. Two objects, A and B, move along the same line. The velocity of A relative to B is

Here $v_A$ and $v_B$ are the velocities measured from the ground. We keep their signs.

Let us try two cases.

• Same direction. A goes at +$20\ \text{m/s}$ and B at +$15\ \text{m/s}$. A relative to B is $20\ \text{m/s} - 15\ \text{m/s} = +5\ \text{m/s}$. A gets ahead of B at $5\ \text{m/s}$.
• Opposite directions. A goes at +$20\ \text{m/s}$ and B at $-15\ \text{m/s}$. A relative to B is $20\ \text{m/s} - (-15\ \text{m/s}) = +35\ \text{m/s}$. They come together at $35\ \text{m/s}$. This is why a train that comes the other way seems so fast.

A third case is the same velocity: A at +$20\ \text{m/s}$ and B at +$20\ \text{m/s}$. Then $20\ \text{m/s} - 20\ \text{m/s} = 0$, and A does not seem to move at all as seen from B.

Whose point of view?

Be careful with the order. $v_{AB}$ is A as seen from B. If we swap them, the sign changes.

If A sees B moving away at $5\ \text{m/s}$, then B sees A moving away at $5\ \text{m/s}$ in the other direction.

You should find that A seems to stand still to B when their velocities are equal, and that it rushes past when they have opposite signs.

To avoid mistakes, always say the sentence aloud: "the velocity of A relative to B". The first letter is the object that you look at. The second letter is the observer.

A moving walkway

A person walks at $1.5\ \text{m/s}$ along a walkway that moves at $2\ \text{m/s}$ in the same direction. From the ground, the person's velocity is $1.5\ \text{m/s} + 2\ \text{m/s} = 3.5\ \text{m/s}$. From the walkway, it is only $1.5\ \text{m/s}$.

To change from one point of view to another, we add or subtract the velocity of the new observer.

Using the rule: let the person be P, the walkway W and the ground G. The person relative to the walkway is $1.5\ \text{m/s}$. The walkway relative to the ground is $2\ \text{m/s}$. So the person relative to the ground is $1.5\ \text{m/s} + 2\ \text{m/s} = 3.5\ \text{m/s}$.

Boats and rivers, and trains

A boat can move at $4\ \text{m/s}$ in still water. A river flows at $1\ \text{m/s}$.

• Downstream, the boat and the river go the same way, so the boat moves at $4\ \text{m/s} + 1\ \text{m/s} = 5\ \text{m/s}$ relative to the bank.
• Upstream, the river works against the boat, so the boat moves at $4\ \text{m/s} - 1\ \text{m/s} = 3\ \text{m/s}$ relative to the bank.

A train example: a train of length $200\ \text{m}$ goes at $20\ \text{m/s}$, and a cyclist rides beside the track in the same direction at $5\ \text{m/s}$. The train relative to the cyclist moves at $20\ \text{m/s} - 5\ \text{m/s} = 15\ \text{m/s}$. To pass the whole cyclist, the train must go $200\ \text{m}$ relative to him, which takes $\dfrac{200\ \text{m}}{15\ \text{m/s}} \approx 13.3\ \text{s}$.

If a cyclist and a train go in opposite directions at the same speeds, the relative speed is $20\ \text{m/s} + 5\ \text{m/s} = 25\ \text{m/s}$, and the train passes in $\dfrac{200\ \text{m}}{25\ \text{m/s}} = 8\ \text{s}$.

Practise with new numbers

Try it in Python

This program finds the velocity of A relative to B for three pairs of objects.

Common mistakes

• Adding when you should subtract. For the same direction, subtract the speeds. For opposite directions, add.
• Forgetting the signs. Choose a positive direction, and put a minus sign on every velocity that goes the other way. Then the formula $v_A - v_B$ always works.
• Mixing the units. Change km/h to m/s before you combine with other data. To change km/h to m/s, multiply by $\dfrac{5}{18}$.
• Mixing up A relative to B with B relative to A. They have opposite signs.

Remember these

• $v_{AB} = v_A - v_B$, with signs.
• Same direction: we subtract the speeds. Opposite directions: we add the speeds.
• The relative velocity is zero when the two velocities are equal.
• $v_{BA} = -v_{AB}$.

Check yourself

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

1. A bus at $54\ \text{km/h}$ passes a cycle at $18\ \text{km/h}$ going the same way. How fast does the cycle seem to move backward, as seen from the bus? Give the answer in m/s.

2. Two trains come towards each other on parallel tracks at $72\ \text{km/h}$ and $54\ \text{km/h}$. How fast do they close in, in m/s?

3. A boat goes at $6\ \text{m/s}$ in still water on a river that flows at $2\ \text{m/s}$. What is its speed relative to the bank, downstream and upstream?

4. A man walks at $2\ \text{m/s}$ along a train that goes at $30\ \text{m/s}$. What is his velocity relative to the ground when he walks to the front, and to the back?

5. Car P sees car Q moving away ahead at $4\ \text{m/s}$. What does Q see P doing?