# Speeding boats

*Speeding Boats printable worksheet*

Two boats travel up and down a lake, each at a constant speed. They turn at each end without slowing down. The boats start at opposite ends of the lake.

If the boats travel at the same speed, whereabouts on the lake will they meet for the first time? What about the second time they meet? Other meetings?

What would happen if one boat travelled twice as fast as the other?

Investigate where the boats will meet for other ratios of speeds.*Once you are confident at working with boats travelling at different speeds, why not try this challenge?*

Can you work out the length of the lake and the ratio of the speed of the boats from the following information?

- The boats leave opposite ends A and B at the same instant
- They cross for the first time 600 metres from A
- After each boat has turned at the end of the lake, they cross for the second time 400 metres from B

Investigate the ratio of the speeds and the length of the lake when you change these distances. Can you find and prove a general result?

This problem is also available in French: Bateaux Sur Lac

Below are two diagrams which show ways of thinking about the problem when one boat travels twice as fast as the other. Can you make sense of them? Do you have your own different way of seeing it?

Uzair from Wilson's School, and Katherine from St Joseph's School both correctly said that at the same speed, boats would always meet in the middle, and if one was going twice as fast as the other they would meet $\frac{2}{3}$ of the way across.

Philip from Wilson's School continued this line of thinking to explain in general where the boats will meet:

When both boats start at opposite ends and travel at constant speed, their first meeting will be halfway down the lake and every meeting thereafter will occur halfway down the lake.

If the ratio of the boats is $2:1$, then boat A is travelling at twice the speed as boat B. They would first meet $\frac{2}{3}$ of the way down the lake. Their second meeting would be at the end of the lake, after boat A has gone two lengths and boat B one.

If the ratio is $3:1$, with boat A going three times as fast as boat B, they would first meet $\frac{3}{4}$ of the way down the lake. Their next meeting would be $\frac{1}{4}$ of the way down the lake, after boat A has gone 2.25 lengths and boat B 0.75 lengths.

If the ratio is $4:1$, with boat A going four times as fast as boat B, they would first meet $\frac{4}{5}$ of the way down the lake.

There is a consistent pattern here: say the amount of times boat A is going faster than boat B is $n$. They would then meet $\frac{n}{n+1}$ of the way down the lake.

Robert clearly explained that boat B must travel further than boat A:

No matter what speed the boats are travelling at, the lake cannot be $600$m or any smaller than that distance. If the lake was smaller than $600$m, the two boats wouldn'??t be able to get $600$m from side A. If the lake was $600$m long, that suggests that boat B would not be moving, however this would be impossible as boat B needs to get $400$m from his starting point for the second crossing.

Boat B must travel further than Boat A. For example, if the lake was $1000$m long, Boat A travels $600$m for the first crossing, and Boat B travels $400$m for the first crossing. Then Boat A has to travel $400$m to the end of the of the lake, and Boat B $600$m to the end of the lake, at this point both boats have travelled $1000$m. Now Boat A has to travel $400$m to the second crossing point $400$m from side B, and Boat B has to travel $600$m to the second crossing point.

So altogether, Boat A has travelled $600 + 400 + 400 = 1400$m and Boat B has travelled $400 + 600 + 600 = 1600$m.

If the lake is $1500$m long, Boat A travels $600$m to the first crossing point, and Boat B travels $900$m to the first crossing point. Next, Boat A has to travel $900$m to the other side of the lake, and Boat B only has to travel $600$m to the other side of the lake. At this point both boats have travelled $1500$m. For the final leg Boat A has to travel $400$m to the second crossing point, and Boat B has to travel $1100$m to the second crossing point.

So altogether, Boat A has travelled $600 + 900 + 400 = 1900$m and Boat B has travelled $900 + 600 + 1100 = 2600$m.

In general, Boat A's distance travelled is Length of lake + $400$m

Boat B's distance travelled is Length of lake $+$ (Length of lake + $400$m).

Extending Robert's ideas:

We know boat B travels further than boat A, so the lake must be more than $1200$m long, as Boat A has travelled $600$m when they meet for the first time.

Let's say the first meeting happens after $1$ unit of time. Then the second meeting is more than $1$ unit of time later as A has to complete the length and turn the corner.

After $3$ units of time, A has travelled $1800$m. If this was the second meeting point, A has travelled Length of lake + $400$m, so the lake must be $1400$m. This would suggest that Boat B travels $800$m in $1$ unit of time, so travels $2400$m in three units of time.

The second meeting point is when Boat B has travelled $1400 + (1400 - 400) = 2400$m, so the solution is that the lake is $1400$m long and Boat A travels at $\frac{3}{4}$ of the speed of Boat B.

### Why do this problem?

This problem offers a chance to think about the links between distance and speed. By solving parts of the problem mentally, there is an opportunity for students to use their visualisation skills and then discuss the different ways of thinking about the problem.

### Possible approach

*This printable worksheet may be useful: Speeding Boats*

Explain the scenario to the class - the boats start from opposite ends and always turn when they get to the other end without slowing down. Ask them to consider whereabouts on the lake the boats will meet for the first time if they travel at the same speed, and where they will meet for the second time, third time and so on. Encourage them to picture it without writing anything down if they can. (It is important to highlight that we're interested in the boats being in the same position on the lake, regardless of which way they are travelling, so a meeting point could be when they cross each other or when one overtakes the other.)

Allow students to share in pairs or small groups their answer to this first question and their different ways of thinking about it. Then set them the task of investigating how the meeting points change if the ratios of the speeds of the boats are changed.

Encourage groups to share their ways of approaching the problem, and give them a chance to describe their way of seeing it. Sharing different strategies for recording can also be useful. Then present the problem of finding the ratio of speeds and length of the lake from the three statements given in the problem. The natural extension to this is to experiment with different distances and see how
it affects the ratio of speeds.

### Key questions

How do you picture what happens for different ratios of speeds?

### Possible support

Circuit Training might be a simpler context to think about as an introduction to this problem.

### Possible extension

Around and Back is a harder problem about ratio of speeds which can be solved using some of the same techniques.