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