Pop Video: A Problem Involving Horses and Carts, and a Fly | Nagwa Pop Video: A Problem Involving Horses and Carts, and a Fly | Nagwa

Pop Video: A Problem Involving Horses and Carts, and a Fly

In this video we look at how a simple maths puzzle can be solved in two different ways, depending upon the point of view you adopt. We also explore the incredible number of assumptions we take for granted when solving such problems.


Video Transcript

A horse and cart sets off down a completely straight track at a speed of three kilometres an hour. At exactly the same time, six kilometres down the road, another horse and cart sets off, heading directly towards the first horse and cart, also at three kilometres an hour. It’s a cart crash waiting to happen.

At the beginning, a small fly takes off from the nose of the first horse and flies in a straight line at a speed of six kilometres an hour toward the nose of the second horse. As soon as it reaches the nose of the second horse, it turns around and flies directly back to the nose of the first horse at a speed of six kilometres an hour. When it reaches the nose of the first horse again, it turns round and heads back to the second horse, and so on.

The pattern is repeated until some time later the horses collide and squash the fly between their noses. How far did the fly travel from the first take-off to the squash incident?

Now this is a problem set in Math World, a slightly strange place in a nearby parallel universe where things work subtly differently than here on Earth. Every math classroom or exam hall doorway is a portal to Math World, as is any piece of paper or screen showing or explaining such a problem.

You’re expected to use the numbers given as an approximate, simple model of a real-world situation. But before we go any further, let’s state a few assumptions that we’d need to make in order to get the answer that the person who set the problem is really looking for.

Assumption number one: The horses and carts are travelling along a completely straight track.

This sounds pretty simple, but there are a couple of possibilities for what is actually going on. Remember, they’re travelling on the surface of the Earth, which is an oblate spheroid with a circumference of between 40008 kilometres pole to pole and 40075 kilometres around the equator.

Of course, they could be travelling east to west or west to east along a line of latitude, which could result in a much smaller radius, depending on how far north or south of the equator they are. Let’s assume they’re travelling around the equator. And we’ll assume a perfectly circular shape, ignoring any of the lumps and bumps and distortions of the Earth’s surface.

If we assume they’re travelling along a six-kilometre arc around the Earth, then the fly will be travelling maybe two metres off the ground, so will be travelling on an arc with a slightly larger radius from the centre of the Earth than that of the ground.

It turns out that the starting points will be nearly 1.9 millimetres further apart at a height of two metres than on the ground. This is negligible over a distance of six kilometres, especially when you consider how difficult it would be to measure the distance between the horses’ noses when they keep moving their heads anyway. We’ll ignore it.

Alternatively, maybe someone has dug out a genuinely level six-kilometre track through the Earth’s crust. If that’s the case, then we can see that the cutting would have a depth of about 70.5 centimetres at its deepest point, which is a relatively minor undertaking in terms of global engineering projects. But it does require work over a six-kilometre stretch of road just for a little mathematical puzzle, so it probably wouldn’t attract the necessary funding. On balance then, it’s okay to operate on a flat Earth basis and assume that we have a straight, level, six-kilometre road.

Assumption number two: The horses and carts travel at three kilometres an hour.

This sounds reasonable. It could mean that they travel at a constant three kilometres an hour, or it could mean that they travel at an average of three kilometres an hour. They’re gonna be travelling for an hour before they collide in the middle. So if the three kilometres an hour is an average speed, any small fluctuations along the way won’t make any difference to the end result, although they could make a huge difference to how you’re able to calculate the answer.

It’s worth just stating for the record that horses in Math World don’t slow down slightly to “let out waste products and exhaust gases” in the way that real-world horses do. So they’re able to maintain a constant speed.

If the speed is a constant three kilometres an hour and the horses started from rest, then it would be a little bit concerning that they were able to instantaneously jerk from rest to three kilometres an hour. This change in velocity in zero time implies infinite acceleration, which under Newton’s second law — the acceleration of an object is dependent upon its mass and the magnitude of an unbalanced external force acting upon it — causes some issues.

Newton’s second law, or “N2” as it’s known to its friends, tells us that the force is equal to the mass times the acceleration. If we say that a horse, small cart, and driver has a mass of around 1000 kilograms, then the forces involved in accelerating this mass instantaneously to three kilometres an hour are essentially infinite and certainly very damaging to the horse, cart, and driver.

It’d be more realistic to arrange the situation so that the horses and carts start a little over six kilometres apart, gently accelerate to a constant three kilometres an hour. And when they’re exactly six kilometres apart, the fly could set off from the first nose to the other nose. This would take very careful coordination and planning but is more feasible than expecting instantaneous achievement of three kilometres an hour for each horse and cart.

For example, someone could stand at the halfway mark and start waving a large flag to signal the start time. The light from the flag would take the same time to reach each cart driver, so they’d start moving at the same time, assuming they’ve got the same reaction times. The horses would start some distance behind their official starting positions, six kilometres apart, and gently accelerate at the same rate, in a well-rehearsed routine, to three kilometres an hour. And we’ll also need to assume that the cart drivers are able to judge speed with great accuracy and control the horses to maintain that exact speed.

Assumption number three: The fly travels at a constant six kilometres an hour backwards and forwards between the noses of the horses.

If we weren’t operating in Math World, there would be a few issues with this aspect of the problem. Firstly, we’d have to train the fly to set off at exactly the right time. This would either be after the infinite force trauma of instantaneously achieving a speed of three kilometres an hour on the first horse’s nose or the fly would need to be capable of judging exactly the right time to set off as the horse whose nose it was sitting on passed the start line.

We’d also have to train the fly to fly backwards and forwards between the two horses’ noses and hope that neither horse breathes in too deeply at an inopportune moment, sniffing up the unfortunate insect to a quite unpleasant mucous-based death up a horse’s nostril.

We’d have to carry out the training without letting slip that the poor fly would be facing certain death at the end of the exercise anyway and hope that its powers of deduction weren’t sufficiently well developed that it would anticipate that the time interval between turnrounds was getting smaller and smaller and at some point there was gonna be a coming-together moment, which would not end profitably for the fly.

I don’t know if you’ve ever watched a fly fly, but travelling in a long straight line doesn’t come naturally to them. They dart around all over the place, in seemingly random directions. So the whole “straight line from nose to nose” piece would need to form a significant part of their training.

I once spent a couple of hours watching a fly dart around my room when I should’ve been doing some homework. It seemed to have an almost inexhaustible appetite for randomly zipping in different directions over short bursts. I didn’t manage to discern what its objectives were, but it never managed to fly in a straight line for more than a metre or two.

This little study episode ended with a power cut, and the lights went out, plunging us into total darkness. The fly was midflight when this happened, and I was very concerned for its welfare, suddenly finding itself flying around without any light, not knowing how or where to land. I listened carefully to see if it bumped into anything, but I never heard it, and I never found out what happened to it next.

As for flying at a constant six kilometres an hour, this may be an even greater challenge. How would it be able to tell that it was travelling at the correct speed? It would be too small to carry a GPS device and certainly not skilled in reading and acting upon such a thing, even if we did have a sufficiently miniaturised piece of equipment.

Then there’s the problem of wind. If the fly maintains a constant air speed, any slight gusts or breeze would mean that its speed relative to the ground would vary. Overcoming this becomes even more complicated if the wind is blowing at an angle to the road, as the fly will have to carefully judge the forward and sideways components of its speed in order to fly exactly along the straight road at six kilometres an hour. Tricky.

Another issue is the instantaneous acceleration from rest to six kilometres an hour at the start and the instantaneous turnaround every time the fly reaches a horse’s nose. We’ve already seen that such magical instant changes of velocity don’t happen in the real world. So we’ll just make an assumption that the acceleration and deceleration phases are exactly balanced by slight changes to the fly’s speed in flight to make sure that its overall average speed is six kilometres an hour.

Lastly, in Math World, we’re able to ignore the fact that six kilometres an hour is close to the maximum speed of around seven kilometres an hour for a regular house fly. So the poor creature is gonna have to fly close to their top speed for an hour without tiring or slowing down. In fact, studies have shown that flies can fly for five or six hours without stopping, so this might not be an unreasonable assumption even in the real world.

Now we’ve made a whole series of observations about how the problem was basically untenable. Let’s make our assumptions and go ahead and find the answer.

I’m guessing that quite a few viewers will already have encountered this problem before and will be aware that there’s a difficult and an easy way to approach the problem. The easy way is to ignore the fly for the moment. Concentrate on the horses. They start six kilometres apart, travel towards each other, both at three kilometres an hour, so they’ll meet in the middle, both having travelled three kilometres. Travelling three kilometres at 3 kilometres an hour takes an hour.

Meanwhile, the fly was travelling back and forth between their noses at a constant six kilometres an hour for the hour. It would have travelled six kilometres. I want to spend a bit of time on the harder solution though, cause it’s a bit more interesting.

If you think about the story from the fly’s point of view, it sets off along a six-kilometre track and encounters a horse’s nose along the way. Then it turns round and flies back the way it came. It travels at six kilometres an hour and soon encounters the same horse’s nose that it started on, although the horse has moved since the start. And it turns round again and pretty soon encounters the second horse again, and so on, and so on.

We end up with a sequence of journeys back and forth, each one getting shorter and shorter, until there is a stromash in the middle, resulting in a squashed fly. Let’s work out how far the fly travels on each leg of its journey.

Let 𝑑𝑓 be the displacement of the fly from its starting point in kilometres. Let 𝑡 be the number of hours that pass. And let 𝑑ℎ two be the displacement of the second horse from the starting point of the first horse and fly, in kilometres.

So 𝑑𝑓 is equal to six times 𝑡, and 𝑑ℎ two is equal to six minus three times 𝑡. Every one hour, the fly travels six kilometres, and the second horse starts with a six-kilometre displacement and reduces this by three kilometres every hour.

Now these equations are simultaneously true. And the fly encounters horse number two when their displacement from the start is the same, so 𝑑𝑓 is equal to 𝑑ℎ two. Then six 𝑡 is equal to six minus three 𝑡. Adding three 𝑡 to both sides means that nine 𝑡 is equal to six. And then dividing both sides by nine, we find that 𝑡 is equal to two-thirds. This means that the fly bumps into the nose of the second horse after two-thirds of an hour, or 40 minutes.

Substituting the time 𝑡 equals two-thirds back into our first equation tells us that the fly travelled four kilometres to encounter the second horse for the first time. Now the fly turns round instantaneously and flies back to the first horse. But remember that both horses travelled two kilometres each in that 40 minutes that the fly was flying between them. And they carry on moving as the fly is flying back to the first horse.

And following the calculations through, we see that the fly travelled four-thirds, or one and one-third, of a kilometre from the second horse’s nose back to the first horse’s nose on that second leg of the journey. We then discover that, for the third leg, it travelled four-ninths of a kilometre. For the next leg of the journey, it was four 27ths, then four 81ths, and so on.

If we look at the series of distances that the fly travels from nose to nose, the terms make a geometric progression. And the first term of that would be four, with a common ratio of a third. It means that each subsequent leg is a third the distance of the previous leg.

And now we just need to add up all those terms. Now the formula that tells us the sum of 𝑛 terms of a geometric progression is the sum of 𝑛 terms is equal to 𝑎, the first term, times one minus 𝑟 the common ratio, to the power of 𝑛, all over one minus 𝑟.

So, for example, by the time the fly has flown back and forth between the horses 14 times, it’s travelled a total of 5.99999875 kilometres. And in fact, at that point, the horses’ noses are just 2.5 millimetres apart. In the real world, the fly would now be squashed. But in Math World, we don’t let minor details like that worry us. We carry on summing an infinite number of terms. And we find out that if we do that, the fly would travel six kilometres exactly.

So in conclusion, if we go along with the Math World assumptions, we’d say that the fly travelled six kilometres. If we made a few concessions to reality, we’d say it flew a couple of millimetres shy of six kilometres, at which time it was squashed. If we try to recreate the situation in the real world, we’d probably say that it’s a ridiculous problem and you couldn’t recreate it.

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