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I was sent some problems by one of the viewers out there.

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I believe his name is Cortagio or Cortajio, I apologize.

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I'm sure I'm mispronouncing it.

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But they are really interesting problems.

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What's interesting about them is that they don't involve

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super-fancy mathematics.

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They just involve an elegant way to apply fairly

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simple mathematics.

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So, without me talking too much about the problems, let's

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try to solve one of them.

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Actually, I don't know where I'm going to categorize this.

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Either in algebra or in the brain-teaser playlist,

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or maybe both.

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So, an officer on horseback starts at the back of a column

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of marching soldiers and rides to the front of the column.

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Then turns around and rides to the rear of the column.

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If the rider travels three times as fast as the column

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moves, and the column is 100 meters long -- OK, I think we

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have enough information to start drawing this.

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Let me draw the column of soldiers.

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I'll just draw this as a big fat line.

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So, that's the column of soldiers.

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I'm about to cough.

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Excuse me.

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I've learned not to cough directly in the microphone.

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Don't want blow your speakers out.

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So that's the column of soldiers right there.

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And the problem tells us that it's 100 meters long.

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So this distance right here is 100 meters.

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And it's moving with some velocity.

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Let's say, to the right.

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So I'm going to just call it, it's moving with

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some velocity, v.

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We'll have to stay abstract there because it doesn't

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tell us the velocity.

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And then we have the officer.

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I wanted to make sure I was giving the correct title.

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So, an officer on horseback stars at the back of a column.

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So he starts here.

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He's on his horse.

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He's on his horse.

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That's my rendition of the horse for these purposes, it

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has an officer on on the back.

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And he is going to, while the whole column is going forward

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with the velocity, v, he's going to go to the front of it.

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Obviously he can go faster because he's on a horse.

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And then he's going to go to the back and they want to

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know how far did this whole column move.

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And how fast is he going?

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Let's see, if the rider travels 3 times as fast as the column,

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if the column is this traveling with the velocity v, he's going

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to be traveling with the velocity of 3v.

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So let's think about two things.

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Let's think about the time it takes him to go to the front

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of the column, and the time it takes to go back.

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So let's say t1 is equal to time to front.

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And then t2 would be time to back.

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There might be other ways to solve this, but this is the

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way that's jumping out into my brain.

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So let's figure out what the time to the front

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of the column is.

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So, over some period of time -- so, how far is the column

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going to move over to t1?

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The column is going to move, well, actually, let me write

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a little formula here.

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Although I'm sure you know this formula.

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Distance is equal to rate times time.

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Right?

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So, over time 1, how far does the column move?

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It's going to move v.

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v times time 1.

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And how far is this guy going to move?

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Well, we're saying over time, when he moves to

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the front of the column.

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So, this whole column is moving to the right.

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And at the same time, this guy's moving to

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the right faster.

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So at the end of time 1, which I've defined as the time

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it takes him to get to the front, what's true about the

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officer on the horseback?

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He will have had to travel 100 meters further than

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the column, right?

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In order to catch up to the front of the column, he would

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have had to go 100 meters further than the column.

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So, the distance that the column travels, plus 100 meters

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is going to be the distance that the officer travels in

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the same amount of time.

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And what's the distance that he travels in that

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same amount of time?

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Well, distance is equal to rate, 3v, times time, t1.

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Time to the front.

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So this is a relationship between velocity and

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the time to the front.

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And let's see if we can simplify this a little bit.

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So, if we track v t1 from both sides, we get 100 is equal to

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3v t1 minus v t1, that's 2v t1, and divide both sides

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by 2, you get 50.

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v t1 is equal to 50.

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The velocity of this column of soldiers times the time it

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takes this officer to get to the front is equal to 50.

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Well, that doesn't solve our problem yet.

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We want to know how far does the column move?

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We have two variables with one equation.

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Not helpful yet.

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Let's see if t2 can help us a little bit?

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All right.

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I'll switch colors to ease the monotony.

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Time to back.

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So, now we're in the opposite situation, where the

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guy's gotten here.

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He turns around.

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I'd argue, immediately.

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He turns around immediately.

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And he goes back.

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With the velocity of 3v.

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So, my question to you as he starts out here, and relative

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to him, he's going this direction at a velocity of 3v.

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And the back of the column is moving towards him with

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the velocity of v, right?

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So if you think about it, the back of the column is going to

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be approaching the rider with the velocity of 4v.

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When you have two velocities that are moving in opposite

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directions -- if I move in this direction, at 60 miles per

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hour, and you're moving in that direction at 60 miles per hour.

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Relative to me, if I assume that I'm stationary, you would

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look like you're coming at me at 120 miles per hour.

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So that same idea -- this officer is going to be

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approaching the end of the column, the back of the column,

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with the velocity of 4v.

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So how long does it take them to get to the back?

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Well, let's see.

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His velocity is 4v.

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I'll do that in green.

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So his velocity is 4v.

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That's how fast he's approaching the back

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of the column.

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And it's going to take him time 2, times time 2.

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And his distance, he's going to travel 100 meters, because

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that's the length of the column.

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Is equal to 100.

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And let's see, if we divide both sides of this by 4,

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we got velocity times time 2 is equal to 25.

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And, once again, we have 1 equation with two unknowns.

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It doesn't help a lot.

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Let's review the problem again to see if somehow we can use

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this information and this information to solve what

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they're asking for.

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So, they want to know how far does the column move by the

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time the officer arrives back at the rear of the column.

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So, they want to know how far did this whole thing move over

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the entire time of this problem happening?

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What was the entire time?

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It was t1 plus t2.

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That's the entire time. t1 to go to the front, and then

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t2 to go back to the back.

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So how far did the column move?

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Well, the column, distance is equal to rate times time.

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So the column will move its distance -- is equal to the

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column's rate, velocity.

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And then what's the time that this whole little

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problem occurs on?

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Well t1 is how long it takes the officer to

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get to the front.

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Plus t2.

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So this is what we're solving for.

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This is what we need to know.

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We need to know the distance traveled by the column.

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And, once again, we have all these variables.

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But maybe we can do something interesting.

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Let's look at this.

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So what is -- so if we just distribute this, we have

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distance is equal to v times t1 plus v times t2.

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And do we know what these things are?

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Well, sure.

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We were able to kind of stumble our way into into what

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these values are.

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And that's what's interesting about this problem.

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We never figured out v.

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We never figured out t1 or t2.

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But we can figure out this whole thing.

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Because v times t1 -- they tell us, is 50.

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So we can substitute it back here.

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So the distance is equal to 50.

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Plus, what's v times t2?

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Well, we solved it here when we figured out how long it would

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take to go back to the front of the column.

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So that's 25.

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So that, we'll put over here.

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So the total distance that the column traveled is 50 plus

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25 meters, or 75 meters.

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And this is a neat problem, because they didn't tell us

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how fast the column is actually going.

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They're just saying it's v.

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They just said that the officer is going 3 times

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as fast, but we don't know absolute velocities.

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But we were still able to figure out, even without even

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knowing the absolute times, we were still able to figure out

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the total distance that the column travelled.

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Anyway, I hope you enjoyed that.

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And if you did, you can thank Cortagio for the problem.

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