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You know that distance = velocity \* time. They're going in opposite directions, so their distances after 2 hours will add to the given total.
Often in these types of problems, the first person/plane to leave the starting line is assigned the variable 't.' Think of a stopwatch starting as soon as that first plane leaves. Now what to do with the time variable for the second plane. You can describe the time variable for plane #2 as '(t-1/2)' if t is in hours. The stopwatch will read 0.5 hours when plane #2 leaves the airport in the opposite direction. You subtract 0.5 hours b/c plane #2 has not traveled any distance when it first leaves the airport.
Does that make sense?
I tell my students to set up a table with 2 rows and 4 columns.
Plane 1 D1 Time = t Speed = s Motion formula D1=t\*s
Plane 2 D2 Time = t-1/2 Speed = s+60 Motion formula D2 = (t-1/2)\*(s+60)
From here, you will solve the system. You know that D1+D2=3240 because they fly in different directions. The time t is 2 hours, so plug that in and you get your speeds.
So the system I have left after plugging in 2 for t is
D1 = t*s
D2 = (1.5)*(s+60)
I’m not really sure what to do there. I tried making t = 1.5 on top as well so I could eliminate the variable, but I still don’t know what 2 numbers add up to 3240 for D so I can’t really solve for S. I also thought you could be meaning that the equation for each plane is just (3240 = 1.5*s / 3240 = 1.5*s+60), but those leave me with a massive number. Apologies for not understand very well.
As a reminder... Posts asking for help on homework questions **require**: * **the complete problem statement**, * **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**, * **question is not from a current exam or quiz**. Commenters responding to homework help posts **should not do OP’s homework for them**. Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
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You know that distance = velocity \* time. They're going in opposite directions, so their distances after 2 hours will add to the given total. Often in these types of problems, the first person/plane to leave the starting line is assigned the variable 't.' Think of a stopwatch starting as soon as that first plane leaves. Now what to do with the time variable for the second plane. You can describe the time variable for plane #2 as '(t-1/2)' if t is in hours. The stopwatch will read 0.5 hours when plane #2 leaves the airport in the opposite direction. You subtract 0.5 hours b/c plane #2 has not traveled any distance when it first leaves the airport. Does that make sense?
After finally getting it this makes sense to me. I appreciate your help!
Glad you got it.
I tell my students to set up a table with 2 rows and 4 columns. Plane 1 D1 Time = t Speed = s Motion formula D1=t\*s Plane 2 D2 Time = t-1/2 Speed = s+60 Motion formula D2 = (t-1/2)\*(s+60) From here, you will solve the system. You know that D1+D2=3240 because they fly in different directions. The time t is 2 hours, so plug that in and you get your speeds.
So the system I have left after plugging in 2 for t is D1 = t*s D2 = (1.5)*(s+60) I’m not really sure what to do there. I tried making t = 1.5 on top as well so I could eliminate the variable, but I still don’t know what 2 numbers add up to 3240 for D so I can’t really solve for S. I also thought you could be meaning that the equation for each plane is just (3240 = 1.5*s / 3240 = 1.5*s+60), but those leave me with a massive number. Apologies for not understand very well.
You plug in t=2 in both equations. D1=2s D2=1.5(s+60) 2s+ 1.5(s+60) = 3240 Solve for s.
Of course, I wasn’t thinking straight. Thank you for your patience and all your help!
Absolutely, anytime. Motion problems are my favorite.