Systems of equations word problems
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Example 4: Solving a word problem with substitution
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Mixture problems 1
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Mixture problems 2
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Mixture problems 3
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Systems and rate problems
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Systems and rate problems 2
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Systems and rate problems 3
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Officer on Horseback
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Two Passing Bicycles Word Problem
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Passed Bike Word Problem
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Passing Trains
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Overtaking Word Problem
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Problem Solving Word Problems 2
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Systems of equations word problems
Passing Trains Trains passing at different rates.
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- 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.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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