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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.