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## Algebra 1

### Unit 4: Lesson 6

Modeling with linear equations and inequalities

# 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.

## Want to join the conversation?

• i don't get it feel so dumb, will see in a few days how it goes • 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. • 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. • 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 • 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.
• 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? • 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)

SET:

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... • 8/2/22
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.

8/3/22
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!

8/4/22
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.  • • 