What seems like a simple question about two trains hurtling towards each other, with a fly in between, actually fools most people. There's an easy way and a hard way of solving this and we run through both below.

**The Problem**

Two trains, on the same track, are 200 miles apart and each is travelling at 50mph towards the other. Starting on one traine is a fly that can fly at 75mph, it flies from one train to the other and back again, until they collide. How far does the fly travel before the trains collide?

Show Solution

**The (Quick) Solution**

The one equation that we need to know for this problem is:
$$ speed = \frac{distance}{time} $$

Let's start by working out how long it is before the two trains collide. They are both travelling towards each other at the same speed, which means that
when they collide they will each have travelled half of the distance separating them, that is half of 200 miles.

We know that each train is travelling at 50mph, so we can work out how long it takes one train to cover its half of the journey:
$$ speed = \frac{distance}{time} $$
$$ \implies 50 = \frac{100}{t} $$
$$ \implies t = \frac{100}{50} = 2 \: \mathrm{hours}$$

We also know that the fly flies at 75mph for 2 hours (the time before the trains collide), so we can work out the distance covered by the fly easily:
$$ speed = \frac{distance}{time} $$
$$ \implies 75 = \frac{d}{2} $$
$$ \implies d = 75 \times 2 = 150 \: \mathrm{miles}$$

That's it! The question can be solved quite simply, but many people don't spot the simple solution, let alone get it correct. There is another way of solving the problem, that is
more difficult...

**The (Slow) Solution**

The other way of calculating how far the fly travels is to work out how far the fly travels on its first leg, before it gets to the other train and turns around, then how far it travels on its second leg and so on.

Let's calculate how far the fly travels on the first leg. We know that the fly and the train it is travelling to must cover 200 miles in total. The fly travels at 75mph
and the train at 50mph, so in total they have a closing speed of 125mph. Therefore the time spent in the first leg is:
$$ speed = \frac{distance}{time} $$
$$ \implies 125 = \frac{200}{t} $$
$$ \implies t = \frac{200}{125} = 1.6 \: \mathrm{hours}$$

During the first leg we can work out how far the fly has travelled as well:
$$ speed = \frac{distance}{time} $$
$$ \implies 75 = \frac{d}{1.6} $$
$$ \implies d = 75 \times 1.6 = 120 \: \mathrm{miles}$$

Next, let's work out how far the fly travels in the second leg. Remember, we know how far the train it has just reached has travelled:
$$ speed = \frac{distance}{time} $$
$$ \implies 50 = \frac{d}{1.6} $$
$$ \implies d = 5 \times 1.6 = 80 \: \mathrm{miles}$$

Therefore the distance the fly and the other train must travel in the second leg is ??200 - 80 - 80 = 40 \mathrm{miles}?? (because both trains have travelled 80 miles). Therefore, at a
closing speed of 125mph this takes:
$$ speed = \frac{distance}{time} $$
$$ \implies 125 = \frac{40}{t} $$
$$ \implies t = \frac{40}{125} = 0.32 \: \mathrm{hours}$$

During the second leg we can work out how far the fly has travelled, as we did before:
$$ speed = \frac{distance}{time} $$
$$ \implies 75 = \frac{d}{0.32} $$
$$ \implies d = 75 \times 0.32 = 24 \: \mathrm{miles}$$

We can repeat the same methodology for the third leg to find that it takes 0.064 hours and the fly travels 4.8 miles. Now, you will notice that
the distances are getting smaller each time and by a constant factor, each subsequent leg has the fly travelling 20% as far as the previous leg. This is a geometric sequence:
$$ 120, \: 24, \:, 4.8, \: ... $$

The sum, ??S?? of a geometric sequence with a start term of ??a?? and a common ration of ??r?? is:
$$ S = \frac{a}{1-r}$$

In this case ??a = 120?? and ??r = 0.2??, so the sum, ??S?? can be found:
$$ S = \frac{a}{1-r}$$
$$ \implies S = \frac{120}{1-0.2}$$
$$ = \frac{120}{0.8}$$
$$ = 150 \: \mathrm{miles}$$

You can see that both approaches give the same answer and although both use relatively simple methods from A-Level Maths
but can take quite different amounts of time to solve in practice. Part of A-Level Maths is knowing which techniques to apply
for which problems, to solve them quickly and without risking making mistakes.