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### Course: Algebra (all content)>Unit 3

Lesson 14: Linear models word problems

# 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 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.
• 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!
• This problem went from 0 to 100 real quick lol
• i don't get it feel so dumb, will see in a few days how it goes
• Trust me, you're not dumb. This is just a little different than what we're used to.
• i usually find myself confortable with sal but this time he went to fast and leave a lot of holes
• 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?
• See other commwnt by Evan Evan or by School920986
• i have never been more confused in my life
• Asslam o alikum. Hi. I too have not understood yet, and I watched this video ,https://youtu.be/OX7ezuU0nzM, the first five minutes of the mister solving the question. Then I rewatched the Khan Academy video and I understood it. I recommend you the video. If you have any questions specifically you can ask me. Have a good day.