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System of equations for passing trains problem

Video transcript
This is the fourth of the five problems that Kortaggio sent me today. And I've been doing these problems because I think they're really neat applications of essentially fairly basic, even you could call it physics, rate problems, and a little bit of algebra. But you're able to figure out pretty neat things, where you're not sure if there's enough information at first. So, here we have train A, represented by these two dots and arrow, is 200 meters long. And I did this because I think we want to be specific about the configurations we're talking to later in the problems. Let's just read it. Train A is 200 meters long. Train B is 400 meters long. They run on parallel tracks at constant speeds. When moving in the same direction, A passes B in 15 seconds. From, so these are the configurations. From A is right behind B, to A is right in front of B. So let me draw that. So, train A, I'll do in this blue color. So train a is like that. And it's 200 meters long. And then train B, I'll do in this green color. Train B is double the length. So it's 400 meters. And so this is a starting configuration. The outline right here. And it says, it takes 15 seconds to pass it, to go to this configuration. So the ending configuration looks like this. Ending configuration. Train B here. And then train A has passed train B. 200 meters. Once again, this is 400 meters here. And this situation takes 15 seconds. So this takes 15 seconds to happen. 15 seconds. So, A passes B in 15 seconds. So this whole sentence right here, is this right here. We go from the situation to that situation. And then in opposite directions, they pass each other in 5 seconds. So from this configuration to that. So let me draw that. So, in the opposite direction example, they start like this. Train A is 200 meters. Train B is facing in the other direction. I'm doing it probably little bit too long, 400 meters. And then in 5 seconds they get from this configuration to this configuration. To that configuration, right there. 200. And, of course, this right here is 400. And this takes-- this right here takes 5 seconds. So in opposite directions they pass each other in 5 seconds. Now, their question is how fast is each train moving in meters per second? So, once again they gave us-- they didn't really give us a lot of velocity information. They just tell us how long it takes to pass. But maybe using both of these pieces of information, we can solve it. So let's say that-- well, you have velocity of train A, so vA. Velocity of train A, and then of course you have the velocity of train B. And regardless of which direction they're facing, it assumes that they're always going at the same velocity. So in this situation, and we always have to just remember, distance is equal to rate times time. So relative, if we assume, because when you take a velocity, you can always take a velocity relative to something else. So, first of all, relative to this train right here, to train B, how far does train A travel? Well, it goes from this point. It goes from right here. To, not just 400 meters, it gets to the point where the front of the train is out here. So it has to travel 400 meters. And another 200 meters. So it travels 600 meters. It travels 600 meters in this situation. And how far does it travel in this situation? Well, once again, if we assume that this train is stationary, and that's, I guess, the key assumption we have to make. We're going to do everything relative. We're going to assume that, even though it is moving at a velocity, the position we're going to make relative. So, relative to this train, we move from right here, we move 400 meters. And then we move 200 more. So, again, in both situations, we move 600 meters. In this situation we move 600 meters, relative. I guess you could say, we move 600 meters relative to the back of train B in this situation. And we move 600 meters relative to the front of train B in this situation. In this situation, we do it in 15 seconds. That's because we're, kind of, where the velocity of train A is being eaten away by the velocity of B. If the velocity of B was 0, if this green train was really stationary, then we'd be moving relative to this train with velocity A. But now this is moving at some velocity. So our relative velocity to this train is going to be something lower. And what is the relative velocity? If you're a passenger sitting in train B, if you're a passenger sitting in train B, right there, how fast will it look like train A is going? What will be the relative velocity? Well, it's going to be the difference between the two. Right? So it's going to be vA minus vB. If you're sitting in this train right there. And you would say, it takes 15 seconds. So, rate times time. Times 15 seconds. Is equal to a distance that it traveled. And, once again, if you're a passenger sitting in this green train right here, you would say, OK, it went from this point. It crossed this entire train then it went another 200 meters. So it went 600 meters. Now, and this is of course in seconds. Now, obviously both of these trains in this have some positive velocity. This train would have moved even more than 600 meters. This train moved 100 and this train would've moved 700 meters. But I'm doing everything relative to what this passenger in the green train sees. Likewise. The passenger in the green train here, let's say the passenger in the green train right here. Oh, I'm doing his arms coming out of his head. But what velocity does he see this blue train coming in at? Well, he's going in this direction at 400 meters per second. The other train is coming in-- no, sorry he's going in this direction at velocity of train B. We don't know what that is. 400 is how long it is. And this train is coming with velocity, want to do it in blue, with velocity a. So you would add the two velocities. If this is coming at 60 miles per hour and this is going in that direction at 60 miles per hour, to this guy who's stationery in train B, he would just feel like this train is approaching him at 120 miles per hour. Or the addition of those two. So from this guy's point of view, this train is approaching with the velocity-- let me do, is approaching with the velocity vA plus vB. And in 5 seconds-- and they give us that information. In 5 seconds-- so velocity times time, or rate times time, is equal to distance-- it travels 600 meters. Remember, this is all relative to the guy, or the gal, sitting on this green train. And that's kind of the key assumption you have to make to make this problem solvable. Well, now we have two equations and two unknowns, we should be able to solve this. For the velocity of the two trains. So, just to simplify, let's divide both sides of this one by 15. So we have the velocity of A minus the velocity of B, is equal to 40 meters per second. Right? 60 divided by 15 is 4. Yep, 40. And then here we have the velocity of A plus the velocity of B, that's an A, is equal to 120 meters per second. And, see, we could just take this equation. Put it down here. We could add the two equations to each other. So if the velocity of A minus the velocity of B is equal to 40 meters per second. Add the two equations. We get 2 times the velocity of A. These two cancel out. Is equal to 160 meters per second. Or the velocity of A is equal to 80 meters per second. And then we can just back-substitute here. The difference between the two is 40. So it's the 80 minus the velocity of B is equal to 40 meters per second. So what's the velocity of B? Well, you could subtract 80 from both sides. You get minus velocity of B is equal to minus 40. Or the velocity of B is equal to 40 meters per second. And we've done the problem. And the key assumption there is to do everything relative to the green guy sitting inside of train B. You could have done it the other way. You could have picked other relative positions. But that, in my brain, is the easiest way to figure it out. So, this guy, in this case, is going at the speed at 40 meters per second. And this guy is going at 80 meters per second. In this situation, this guy is traveling at 80 meters per second and this guy going in this direction at 40 meters per second. Anyway. Thanks again to Kortaggio for that problem.