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Comparing linear rates example

Compare the positions of two creatures moving at constant speed and determine when one catches up with the other. Created by Sal Khan.

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  • blobby green style avatar for user Medic7443
    I am so glad I am not the only one lost here. We went from nice simple equations and inequalities to this complex mess in like 2 seconds. This video offers little to no explanation and simply gives you a long mess of how to solve one issue. So unless I need to figure out how far my dragon and griffin have flown this will never help me in life.
    (81 votes)
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    • piceratops ultimate style avatar for user elijah
      This actually can help you in life! If you are calculating two things flying (Birds, drones, planes, ETC), over a space (A house, a field, a runway), this can help you figure out the same type of equation, but this video does not do the best job explaining how to solve this. Good luck and hope this helps!
      (11 votes)
  • blobby green style avatar for user delisespinal
    This problem went from 0 to 100 real quick lol
    (75 votes)
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  • blobby green style avatar for user lalo graza
    i usually find myself confortable with sal but this time he went to fast and leave a lot of holes
    (48 votes)
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    • hopper cool style avatar for user MaeS
      I know, I understand. He went from being easy to understand to suddenly talking in Alien or something. I was very confused, so don't worry.
      You should probably just serach it up on Google or something, it might have the answer and explanation you need.
      (1 vote)
  • leaf blue style avatar for user edc
    i don't get it feel so dumb, will see in a few days how it goes
    (41 votes)
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  • blobby green style avatar for user user
    This video makes 0 sense i dont get how to do this equation which means i cant do any of the problems and its driving me insane. How do i even know what t is and why is it 50x(t+42) how and why. WHAT IS A T KILOMETER. Why isnt he just dividing 60/175 then x35 hes making it more complicated and making other problems more difficult. What is a more simple way to do this stupid problem none of what he doing is registering. OR at least explain everything in more simple terms
    (20 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      Well, 60∕175⋅35 definitely gives us the correct answer.
      The question is why it gives us the correct answer.

      – – –

      1 minute = 1∕60 of an hour.
      Thereby 42 minutes (the time the gryphon spent flying away from the castle until the dragon arrived at the castle) = 42∕60 = 0.7 hours.

      Flying at a speed of 50 km∕h for 0.7 hours, the gryphon would then be
      50⋅0.7 = 35 km away from the castle when the dragon arrived at the castle.

      Now, instead of having the gryphon continue flying at a speed of 50 km/h and having the dragon pursue the gryphon at a speed of 225 km/h,
      we realize that it would take the exact same amount of time for the dragon to catch up with the gryphon if the gryphon was sitting still (35 km away from the castle) and the dragon was flying at a speed of 225 − 50 = 175 km/hour.

      Thus, all we need to calculate is how long it would take the dragon to fly 35 km at a speed of 175 km/h,
      which would be 35∕175 hours.

      1 hour = 60 minutes,
      so 35∕175 hours = 35∕175⋅60 minutes = 12 minutes.

      – – –

      I don't know if this solution is any easier to follow along with than the solution Sal presented in the video, but it is at least equally valid.
      (43 votes)
  • sneak peak green style avatar for user Robert Zak
    I've got no idea of what I just watched, didn't understand anything.

    Are you sure this lecture is positioned correctly within the curriculum? Because we went jumped from simple equations to something like this?
    (19 votes)
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  • blobby green style avatar for user atomluna00
    i have never been more confused in my life
    (15 votes)
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  • duskpin sapling style avatar for user somebody
    This video is a complete disappointment. After watching it, I still neither know why the equation to solve the problem is set up as it is nor how to solve these kinds of problems myself and I'm sure that lots of students who have written similar comments would agree. I was looking for a general method to be able to solve these kinds of questions and now I am even more confused.
    (17 votes)
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  • male robot hal style avatar for user Pj Kelley
    PLEASE. Re-do this video but with a clearer explanation. I love the videos so far, but this was just not very well explained
    (16 votes)
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  • blobby green style avatar for user Cara Horner
    i had to watch this video 3 times to understand this. i feel like this is way too complicated for algebra 1. i mean we went from plotting lines, to this?? Slow ya roll khan academy...
    (15 votes)
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Video transcript

- [Instructor] We're told that a gryphon flew east over a castle at 50 kilometers per hour. Then, 42 minutes later, a dragon also flew east over the castle. The dragon flew 225 kilometers per hour. Assume both the gryphon and the dragon continue flying east at the same speeds. How many minutes will the dragon have flown since passing the castle when it catches up to the gryphon? They also ask us how many kilometers east of the castle will they be at that time? So pause this video and see if you can figure this out before we do this together. All right, so the question is, how many minutes will the dragon have flown since passing the castle when it catches up to the gryphon? So let's set that variable to be equal to t, the number of minutes that the dragon has flown, dragon flown since castle, since castle and catches up, catches up. So let's think about the distance that the dragon would have flown in that t minutes. Well, the dragon's flying at 225 kilometers per hour. So the distance is going to be the rate, 225 kilometers per hour, times the time, so times t minutes. But we have to be careful. This is in minutes, while the rate is given in kilometers per hour. So we have to make sure that our units work out. And so for every one hour, we have 60 minutes. And we can see here that the units, indeed, do work out. This hour cancels with that hour in the numerator and the denominator, and this minutes cancels out with this minutes. And so the distance that the dragon would have flown after t minutes is going to be 225t over 60 kilometers. So let me write it this way. 225 over 60t kilometers. Now we could try to simplify this, but I'll leave it like this for now. Maybe I'll simplify it a little bit later. Now let's think about how far the gryphon would have flown. So they tell us that the gryphon is flying at 50 kilometers per hour, so 50 kilometers per hour. And how long would the gryphon have flown by that point? Well, the gryphon passed the castle 42 minutes before the dragon passed it. So if t is how many minutes that the dragon has been flying east of the castle, well, then the gryphon is going to be t plus 42 minutes. So t plus 42 minutes is how long that the gryphon has been traveling east of the castle. And then once again, we have to make sure that our units work out. So we're gonna say one hour for every 60 minutes. The minutes cancel out, the hours cancel out, and so we are going to be left with 50 over 60, or I could write 5/6 times t plus 42 kilometers. Or if we wanna simplify this even more, this is going to be 5/6 t plus, let's see, 5/6 of 42, 42 divided by 6 is 7, times 5 is 35, plus 35 kilometers. So we know that they would have flown the exact same distance because we're talking about when the dragon catches up with the gryphon. So these two things need to be equal to each other, and then we can just solve for t. So let's do that. We get to 225 over 60t, and we know that both sides are in kilometers, so I, just for the sake of simplicity, I won't write the units here. So this is going to be equal to 5/6 t plus 35. And now let us solve for t. We can subtract 5/6 t from both sides, or actually, since I already have 60 as a denominator, I could subtract 50 over 60 t from both sides, which is the same thing as 5/6 t. So I am going to have 225 over 60 minus 50 over 60. And then all of that times t is equal to 35. And so let me get myself a little bit more real estate. So this is going to be simplified as 175 over 60 t is equal to 35. Or, then if I just multiplied both sides by 60 over 175, I will get the t is equal to 35 times 60 over 175. And you might recognize that 35 is the same thing as 5 times 7, and 175 is the same thing as 25 times 7. So these sevens cancel out. And then if we divide both this and this by 5, this becomes a 1, this becomes a 5. And then, 60 divided by 5 is equal to 12. And so remember, t was in minutes. So the answer to the first part of the question is 12 minutes. So let's go back up to what they were asking us. How many minutes will the dragon have flown since passing the castle when it catches up to the gryphon? Well, we defined that as t, and then we got 12 minutes. Now, the next part of the question is how many kilometers east of the castle will they be at that time? So to figure out how many kilometers east of the castle, we have to calculate this expression or this expression when t is equal to 12. So I'm going to use this first one. So you're going to have 225 over 60 times 12 is going to give us, let's see, 60 and 12 are both divisible by 12, so you get that 1 over 5. And if you divide 225 by 5, that is going to give us 45, and the units all work out to kilometers. So we answered the first two parts of the question. And the second part of this question, they tell us that Latanya and Jair both wrote correct inequalities for the times, in minutes, when the dragon is farther east of the castle than the gryphon is. Latanya wrote t is greater than 12, and Jair wrote t is greater than 54. How did Latanya and Jair define their variables? Well, Latanya defined her variables the exact same way that I defined mine, because we got t is equal to 12 when the dragon passes up the gryphon. So for t is greater than 12, the dragon is farther east of the castle than the gryphon. So we did the exact same thing as Latanya. But what Jair did, if he got t is greater than 54, is he must have defined t as being equal to the number of minutes since the gryphon, the slower, the first but slower creature, passed the castle. And so he would have gotten t is greater than 54. And then if you wanted to know how many minutes since the dragon passed the castle, because the dragon got there 42 minutes later, he would have subtracted 42 from that. And that's how you connect these two numbers over here. But the important thing to realize is there's multiple ways to solve the same problem. What matters is to be very clear, how you are defining that variable and use it consistently throughout, and then interpret it correctly when you're answering the questions.