Systems of equations word problems
-
Example 4: Solving a word problem with substitution
-
Mixture problems 1
-
Mixture problems 2
-
Mixture problems 3
-
Systems and rate problems
-
Systems and rate problems 2
-
Systems and rate problems 3
-
Officer on Horseback
-
Two Passing Bicycles Word Problem
-
Passed Bike Word Problem
-
Passing Trains
-
Overtaking Word Problem
-
Problem Solving Word Problems 2
-
Systems of equations word problems
Overtaking Word Problem 3 people overtaking each other at different rates
⇐ Use this menu to view and help create subtitles for this video in many different languages.
You'll probably want to hide YouTube's captions if using these subtitles.
- Here's another problem that Kortaggio sent me and,
- like the other two, this is quite interesting.
- Although, I think you'll start to see a pattern forming
- in how these are solved.
- And that's the whole point, not just to show you neat problems,
- but also to give you an intuition of how to solve it.
- So what does it say?
- It says Alice is 100 meters from Bill and Bill is 300
- meters away from Chelsea.
- Maybe this is a bit of a Clintonian reference, but who
- knows, or maybe they just wanted people with starting
- names a, b, and c.
- So let's draw that out.
- So Alice is 100 meters from Bill-- all right.
- Alice is 100 meters from Bill-- and actually he drew a little
- diagram, so we know that also Chelsea is to the
- right of Bill.
- So let's see, the diagram he sent me looks like this.
- So you have Alice.
- And then 100 meters away you have Bill.
- And then 300 meters to the right of that you have Chelsea.
- This distance right here is 100 meters.
- And then this distance right there is 300 meters.
- And the problem tells us they are all facing east and are
- standing on the same line.
- Fair enough.
- We'll say that's east.
- So they're all facing in that direction standing
- in the same line.
- They all travel to the east at constant speeds.
- OK, but they don't say that they're all traveling
- at the same speed.
- Each individual speed is constant.
- So this could be velocity of Alice, lowercase a, velocity
- of Bill, and then this is the velocity of Chelsea.
- Alright, they all travel east at constant speeds.
- In 6 minutes, Alice overtakes Bill.
- So what has to happen?
- Let me do this in another color and write down what that
- equation would look like.
- So I'll do it in green.
- In 6 minutes, Alice overtakes Bill.
- This tells the distance that Alice travels in 6 minutes.
- So what's the distance that Alice travels in 6 minutes?
- It would be the velocity of Alice times 6.
- And she overtakes Bill.
- So that means she goes 100 meters more than Bill went.
- So that means she went 100 meters plus the distance that
- Bill went in that 6 minutes.
- And the distance that Bill went in 6 minutes is
- velocity of Bill times 6.
- Right?
- This tell us in 6 minutes, Alice went 100 meters more
- than however far Bill went in 6 minutes.
- That's that equation there.
- We got it from this information.
- And then the next thing, this problem tells us, in
- another 6 minutes Alice overtakes Chelsea.
- So from the beginning of our time, it essentially
- takes 12 minutes.
- Right?
- 6 minutes to overtake Bill and then another 6 minutes
- to overtake Chelsea.
- So 12 minutes to overtake Chelsea.
- So we could write that as the distance that Alice travels
- in 12 minutes is equal to--
- Well, she has to overtake Chelsea.
- That means she made up 400 meters of distance, so that
- she traveled 400 more meters.
- So the distance that Alice travels is going to be 400
- meters more than the distance that Chelsea travels
- in 12 minutes.
- And remember, distance is just equal to rate times time.
- Right?
- So in 12 minutes this is how far-- oh sorry --this is how
- far Chelsea travels, that's a c.
- And this is how far Alice travels, and she is going
- to travel 400 meters more.
- OK, that's what those two equations tell us.
- And what is their question?
- They're asking us, how many minutes did it take Bill
- to overtake Chelsea?
- How many minutes did it take Bill to overtake Chelsea?
- All right, so let's say we want to know time in minutes.
- How long does it take Bill to overtake Chelsea?
- Let's say we know that it's time, t.
- So that tells us that the velocity of Bill times this
- time is going to be equal to --for him to overtake Chelsea,
- he has to travel 300 meters more than her in that
- same amount of time.
- So he has to go 300 meters more than how far Chelsea
- travels in that time.
- So that's what this is, and then we essentially are just
- trying to solve for this time.
- We want to be able to solve for this.
- Let's solve this equation.
- Remember, all these equations really have the same form.
- In order for this person to overtake this
- person in 6 minutes.
- That means in 6 minutes she would have to travel 100
- meters more than him.
- In 12 minutes she would have to travel 400 meters
- more than Chelsea.
- And then in t minutes-- and that's what we're going to have
- to solve the problem for --b is going to have to travel
- 300 meters more than Chelsea right there.
- We just solve for t, so let's just solve for t.
- So what do you get?
- You get vbt minus vct is equal to 300.
- I just subtracted Chelsea's velocity times time from
- both sides of the equation.
- We can factor out the time, so let's do that.
- So you get vb minus vc times time-- I'm just factoring out
- the time --is equal to 300.
- Or time is equal to 300/vb minus vc.
- So if we could somehow figure out what this is equal
- to, then we would have solved our problem.
- Let's see if we can use the other information.
- Let's see, if we solve for vb here and solved for vc here,
- hopefully the va's will cancel out.
- And I suspect they will, otherwise this problem
- would not be solvable.
- So let's do that.
- Let's rewrite this equation up here.
- Let's solve for vb, for Bill's velocity.
- Let's see, if we divide both sides by 6-- well, let's
- subtract 100 from both sides.
- So you have Bill's velocity times 6-- that's just that
- --is equal to 6 times Alice's velocity minus 100.
- All I did is I subtracted 100 from both sides of
- this equation and I swapped the sides.
- Divide both sides by 6.
- You got the velocity of Bill is equal to the velocity
- of Alice minus 100/6.
- 100/6 or 50/3-- I'll just keep it as a 100/6.
- Fair enough.
- Let's see if we can do the same thing with this equation.
- If we subtract 400 from both sides, we get 12 times
- Chelsea's velocity is equal to 12 times Alice's
- velocity minus 400.
- Divide both sides of this equation by 12.
- You get Chelsea's velocity is equal to Alice's
- velocity minus 400/12.
- All right.
- So let's if we can substitute this back in for vb, and
- substitute this back in for vc.
- And hopefully the va's cancel out.
- It already looks like they should.
- So our time, the time that it takes for Bill to overtake
- Chelsea-- remember that's what t was, it's always good to
- remind yourself what this whole problem was about to begin
- with --is equal to 300/vb.
- Well instead of vb let's write this.
- Va, the velocity of Alice, minus 100/6 minus vc.
- So minus this right here.
- Minus va minus 400/12.
- This should simplify to t-- I'll just arbitrarily
- switch colors.
- t is equal to 300/va minus-- what's 100/6?
- 100/6 is the same thing as 50/3.
- Distribute the minus sign.
- Minus va-- we can already see these are going to cancel out
- --minus times a minus plus 400/12 is the same thing
- is 200/6 or 100/3.
- Right?
- Plus 100/3.
- These two cancel out, va minus va.
- So then we're left with t is equal to 300 over
- 100/3 minus 50/3.
- Right?
- Or I could say minus 50/3 plus 100/3, but either way, minus
- 50/3 plus 100/3 is just 50/3.
- Did I do that right?
- If I divide the top and the bottom by 4, I get this.
- And then sure, this is a 100/3 and than 50/3.
- And so when you divide by a fraction that's the same thing
- is multiplying by its inverse.
- 300 times 3/50 over 1.
- And then we can cancel out some terms, so we don't have
- to multiply big numbers.
- So if we divide the top and the bottom by 50 that becomes 1,
- this becomes 6, and then we're left with t is equal to 6 times
- 3/1, which is 18, and all the time units we were working with
- the whole time were in minutes.
- And we're done.
- That was a little hairier than the other two, but notice we
- were able to solve it by just algebraically writing down the
- information that they actually gave us in the problem and just
- seeing what we have to solve for and then substituting back
- in and then miraculously things canceled out.
- And in general, if you're given a nice problem like this you
- can normally just, you know, march forward knowing that if
- you do things correctly probably nice things will
- happen and terms will cancel out.
- But I thought you would find that interesting and once
- again, thanks to Kortaggio for that problem.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
|
Have something that's not a question about this content? |
This discussion area is not meant for answering homework questions.
Discuss the site
For general discussions about Khan Academy, visit our Reddit discussion page.
Flag inappropriate posts
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
abuse
- disrespectful or offensive
- an advertisement
not helpful
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
wrong category
- a tip or feedback in Questions
- a question in Tips & Feedback
- an answer that should be its own question
about the site
Share a tip
Suggest a fix
Have something that's not a tip or feedback about this content?
This discussion area is not meant for answering homework questions.