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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- 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.(16 votes)
- If you are having an extremely hard time with this problem in Algebra 1, don't stress. This problem is too advanced for what you've learned so far. It has some new concepts that aren't explained. Plus, it is further complicated by requiring conversion between minutes and hours. There really should be more lessons before this.
Here is what I suggest you do. Try the practice quiz that comes after this, get them wrong and look at the answers. The answers give a very clear step by step, which IMO explain it much, much better than this video.
And if that doesn't help, here is another thing you can try. This concept is basically the same thing as what you will learn in the next two units of Algebra 1. Those units are "Forms of Linear Equations" and "Systems of Equations". So you can also try skipping ahead and doing the next 2 units, then coming back here and trying this again.(15 votes)
- 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(9 votes)
- 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.(7 votes)
- 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?(10 votes)
- I will try to explain the equation because the video was kind of confusing...
We have a gryphon that is going 50 km/h
AND a dragon that is going 225 km/h but started 42 mins later.
We first have to convert hours to mins
(50 km/h = 50 km/60 mins and 225 km/h = 50 km/60 mins)
50/60 (t+42) = 225/60 (t)
Why is t+42 used? This is because t is the amount of time that passes until the dragon meets the gryphon. The gryphon leaves 42 minutes earlier, so it travels for t+42 minutes.
In a way, this is a demonstration of d = rt
The distances that are travelled must be equal when:
distance gryphon travels = distance dragons travels
50/60 * (t+42) = 225/60 * t
rate * time = rate * t
I really hope this helps...(8 votes)
This is likely a very dumb question, but I'm kind of stumped so I have to ask it.
I've spent two evenings on the problem in this video and am still confused. Maybe I shouldn't have paused the video to see if I could figure it out for myself, but I did and I couldn't. It looked something like a system of equations to me, but I couldn't figure out the equations. I tried to graph it but failed. I told myself my error had been to have the dragon starting out at t=42 and they both had to start at 0, that the point when they passed over the castle was the only thing that mattered, but tonight that didn't work out and the only way I succeeded in graphing it was indeed to put the dragon's start at t=42. The gryphon's distance was easy, 5/6·t, but the dragon's was more difficult. I made a table with points (42,0), (44,7.5), (46,15), (50,30) and (54,45), and tried to make the formula d=3.75·(t+42) out of that. It didn't work, though: it had to be 3.75 times quantity t MINUS 42. This is what I can't get my head around. Why is the formula, independent of the gryphon's formula, 3.75 times t MINUS 42 when we're over to the right of the graph at 42-54, in territory that I'd think would be plus? I would like not to spend a third evening freaking out about this, but the way this is(n't) going it looks like this may happen. Thanks.
I've gotten past the problem with plus vs. minus, realizing that it's not a matter of the distance between 0 and 42 ("plus"), but between t and 42 ("minus", with t greater than 42 except when y equals 0). Moreover, in addition to graphing the solution – from origins (0,0) and (42,0) – I have solved the problem as a system of equations – y=(5/6)x and y=(15/4)x-157.5) – by both substitution and elimination. But all this wound up taking me three days with all my fuzziness, misconceptions and arithmetical errors, while Sal's solution is much simpler, not explicitly involving a system of equations and not involving a y-intercept as in the formula for the dragon that I worked out. The y-intercept also seems practically irrelevant in this case, as we don't know whether or how long the dragon flew in a straight line due east before it passed over the castle.
So my question now is, how do I get from my apparently overly complicated solution to Sal's relatively simple one? Something seems to be missing, and maybe it's a general equation for a proportional relation with an x-offset – in other words, a d=rt with a modified t. Or maybe I'm still thinking wrong, as Sal seems to be doing something different in holding the identical t variable through both equations rather than solving for it separately as in the system of equations solutions. If anyone understands my confusion here I hope they can help in resolving it. Thanks!
I found the "proportional relation with an x-offset", the "d=rt with a modified t" I was looking for. It's just the point-slope formula, which I'd forgotten while obsessing on slope-intercept.
y - y_0 = m (x - x_0)
With (42,0) as (x_0,y_0)
y - 0 = m (x - 42)
y = m (x - 42)
which is the formula I got previously, so I'm finally content with it. I don't really need to get from any hard-fought solution of mine to Sal's simple intuitive one with a single t variable, so I'm going to finally leave this video behind and move on.(4 votes)
- Usually we would say that the distance covered by the dragon is 𝐷(𝑡) = 3.75𝑡 .
But with 𝑡 being the time that's passed since the gryphon flew over the castle,
we know that at 𝑡 = 42 the dragon's distance from the castle is 0, which is why you would subtract 42 from 𝑡:
𝐷(𝑡) = 3.75(𝑡 − 42) ⇒ 𝐷(42) = 0(3 votes)
- Dude, there is an easier way to do this. First, the griffin started 42 minutes earlier so 42/60x50 and we would get 35. so the griffin traveled 35 miles before the dragon began flying after it. We know that 225-50 is 175. 35/175=1/5 so it took 1/5 hours for the dragon to catch up to the griffin(4 votes)
- where do they get the 35 from(2 votes)
- [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.