Meeting and catching up

The question of every chase

A police car starts to follow a speeding motorbike. When does it catch the motorbike? How far have they both gone by then?

To answer, we write down where each one is at any time $t$. Then we find the time when the two positions are the same.

The method

1. Choose a starting point (position zero) and a positive direction.
2. Write the position of each object at time $t$. For a steady speed, $x = x_0 + vt$, where $x_0$ is where it starts.
3. Put the two positions equal, and solve for $t$.
4. Put this $t$ back into either position to find where they meet.

An example with a head start

Car A leaves first at $15\ \text{m/s}$. Car B leaves $5\ \text{s}$ later from the same place, at $20\ \text{m/s}$. When does B catch A?

We measure the time from the moment A leaves. At time $t$:

• A is at $15t$ metres, with $t$ in seconds.
• B has been driving for $t - 5$ seconds, so it is at $20(t - 5)$.

They meet when the two positions are equal.

They meet $20\ \text{s}$ after A left, at $15\ \text{m/s} \times 20\ \text{s} = 300\ \text{m}$ from the start.

On the graph, the two lines cross at the meeting point. Before that, A is ahead. After that, B is ahead.

You should find that a longer head start moves the meeting point later and farther, and that the lines never cross when B is not faster than A.

The same answer with relative velocity

There is a shorter way. When B starts, A is already $15\ \text{m/s} \times 5\ \text{s} = 75\ \text{m}$ ahead. B closes this gap at $20\ \text{m/s} - 15\ \text{m/s} = 5\ \text{m/s}$. So B needs $\dfrac{75\ \text{m}}{5\ \text{m/s}} = 15\ \text{s}$ after it starts. That is $20\ \text{s}$ after A left. It is the same answer.

Moving towards each other

Two objects can also come towards each other. They meet when the distances that they cover add up to the gap between them. Their closing speed is the sum of their speeds.

A gap of $600\ \text{m}$ closes at $20\ \text{m/s} + 10\ \text{m/s} = 30\ \text{m/s}$. So they meet after $\dfrac{600\ \text{m}}{30\ \text{m/s}} = 20\ \text{s}$.

Practise with new numbers

Try it in Python

This program shows the position of both cars every 5 seconds. Then it finds the time when they meet.

Remember these

• Same direction, one catches up: closing speed = difference of the speeds.
• Opposite directions, coming together: closing speed = sum of the speeds.
• Time to meet = gap ÷ closing speed.
• Always check that both objects have started. A car that has not started cannot close any gap.

Check yourself

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

1. A truck goes at $10\ \text{m/s}$. A car $100\ \text{m}$ behind it goes at $15\ \text{m/s}$. When does the car reach the truck?

2. Two runners, $300\ \text{m}$ apart, run towards each other at $4\ \text{m/s}$ and $6\ \text{m/s}$. When do they meet?

3. A bus leaves a stop at $8\ \text{m/s}$. A boy $40\ \text{m}$ behind runs after it at $5\ \text{m/s}$. Does he catch it?

4. On a position-time graph, what does the point where two lines cross tell you?