Transcription
I was sent some problems by one of the viewers out there. I believe his name is Cortagio or Cortajio; I apologize. I'm sure I'm mispronouncing it. But they are really interesting problems. What's interesting about them is that they don't involve super-fancy mathematics. They just involve an elegant way to apply fairly simple mathematics. So, without me talking too much about the problems, let's try to solve one of them.
Actually, I don't know where I'm going to categorize this. Either in algebra or in the brain-teaser playlist, or maybe both. So, an officer on horseback starts at the back of a column of marching soldiers and rides to the front of the column. Then he turns around and rides to the rear of the column. If the rider travels three times as fast as the column moves, and the column is 100 meters long -- OK, I think we have enough information to start drawing this. Let me draw the column of soldiers. I'll just draw this as a big fat line. So, that's the column of soldiers. I'm about to cough. Excuse me. I've learned not to cough directly in the microphone. Don't want to blow your speakers out. So that's the column of soldiers right there. And the problem tells us that it's 100 meters long. So this distance right here is 100 meters. And it's moving with some velocity. Let's say, to the right. So I'm going to just call it, it's moving with some velocity, v. We'll have to stay abstract there because it doesn't tell us the velocity.
And then we have the officer. I wanted to make sure I was giving the correct title. So, an officer on horseback starts at the back of a column. So he starts here. He's on his horse. He's on his horse. That's my rendition of the horse for these purposes; it has an officer on the back. And he is going to, while the whole column is going forward with the velocity, v, he's going to go to the front of it. Obviously, he can go faster because he's on a horse. And then he's going to go to the back, and they want to know how far did this whole column move. And how fast is he going? Let's see, if the rider travels 3 times as fast as the column, if the column is traveling with the velocity v, he's going to be traveling with the velocity of 3v.
So let's think about two things. Let's think about the time it takes him to go to the front of the column, and the time it takes to go back. So let's say t1 is equal to time to front. And then t2 would be time to back. There might be other ways to solve this, but this is the way that's jumping out into my brain. So let's figure out what the time to the front of the column is. So, over some period of time -- so, how far is the column going to move over t1? The column is going to move, well, actually, let me write a little formula here. Although I'm sure you know this formula. Distance is equal to rate times time. Right? So, over time 1, how far does the column move? It's going to move v times time 1. And how far is this guy going to move? Well, we're saying over time, when he moves to the front of the column. So, this whole column is moving to the right. And at the same time, this guy's moving to the right faster. So at the end of time 1, which I've defined as the time it takes him to get to the front, what's true about the officer on the horseback? He will have had to travel 100 meters further than the column, right? In order to catch up to the front of the column, he would have had to go 100 meters further than the column. So, the distance that the column travels, plus 100 meters is going to be the distance that the officer travels in the same amount of time. And what's the distance that he travels in that same amount of time? Well, distance is equal to rate, 3v, times time, t1. Time to the front. So this is a relationship between velocity and the time to the front. And let's see if we can simplify this a little bit. So, if we subtract v t1 from both sides, we get 100 is equal to 3v t1 minus v t1, that's 2v t1, and divide both sides by 2, you get 50. v t1 is equal to 50. The velocity of this column of soldiers times the time it takes this officer to get to the front is equal to 50. Well, that doesn't solve our problem yet. We want to know how far does the column move? We have two variables with one equation. Not helpful yet.
Let's see if t2 can help us a little bit? All right. I'll switch colors to ease the monotony. Time to back. So, now we're in the opposite situation, where the guy's gotten here. He turns around. I'd argue, immediately. He turns around immediately. And he goes back with the velocity of 3v. So, my question to you is, as he starts out here, and relative to him, he's going this direction at a velocity of 3v. And the back of the column is moving towards him with the velocity of v, right? So if you think about it, the back of the column is going to be approaching the rider with the velocity of 4v. When you have two velocities that are moving in opposite directions -- if I move in this direction, at 60 miles per hour, and you're moving in that direction at 60 miles per hour. Relative to me, if I assume that I'm stationary, you would look like you're coming at me at 120 miles per hour. So that same idea -- this officer is going to be approaching the end of the column, the back of the column, with the velocity of 4v. So how long does it take them to get to the back? Well, let's see. His velocity is 4v. I'll do that in green. So his velocity is 4v. That's how fast he's approaching the back of the column. And it's going to take him time 2, times time 2. And his distance, he's going to travel 100 meters, because that's the length of the column. Is equal to 100. And let's see, if we divide both sides of this by 4, we got velocity times time 2 is equal to 25. And, once again, we have 1 equation with two unknowns. It doesn't help a lot.
Let's review the problem again to see if somehow we can use this information and this information to solve what they're asking for. So, they want to know how far does the column move by the time the officer arrives back at the rear of the column. So, they want to know how far did this whole thing move over the entire time of this problem happening? What was the entire time? It was t1 plus t2. That's the entire time. t1 to go to the front, and then t2 to go back to the back. So how far did the column move? Well, the column, distance is equal to rate times time. So the column will move its distance -- is equal to the column's rate, velocity. And then what's the time that this whole little problem occurs on? Well t1 is how long it takes the officer to get to the front, plus t2. So this is what we're solving for. This is what we need to know. We need to know the distance traveled by the column. And, once again, we have all these variables. But maybe we can do something interesting. Let's look at this. So what is -- so if we just distribute this, we have distance is equal to v times t1 plus v times t2. And do we know what these things are? Well, sure. We were able to kind of stumble our way into into what these values are. And that's what's interesting about this problem. We never figured out v. We never figured out t1 or t2. But we can figure out this whole thing. Because v times t1 -- they tell us, is 50. So we can substitute it back here. So the distance is equal to 50. Plus, what's v times t2? Well, we solved it here when we figured out how long it would take to go back to the front of the column. So that's 25. So that, we'll put over here. So the total distance that the column traveled is 50 plus 25 meters, or 75 meters. And this is a neat problem, because they didn't tell us how fast the column is actually going. They're just saying it's v. They just said that the officer is going 3 times as fast, but we don't know absolute velocities. But we were still able to figure out, even without even knowing the absolute times, we were still able to figure out the total distance that the column traveled. Anyway, I hope you enjoyed that. And if you did, you can thank Cortagio for the problem.