anonymous
  • anonymous
HELP!! How to set this up? you comment 56 miles one way to work. The trip to work takes 10mins linger than the trip home. YOu average speed on the trip home is 8 miles faster. what is your average speed on the trip home?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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phi
  • phi
Comute (not comment) 56 miles one way. Rate * time = distance sounds useful we know a 1-way trip distance is 56, we we know rate * time= 56 Let's call the rate to work S (for speed) in miles per hour and the time T (in hours)
phi
  • phi
with the definitions rate to work S (for speed) in miles per hour and the time T (in hours) you get the equation S*T=56 (going to work) now we need an equation for going home. trip to work takes 10mins longer than the trip home and hopefully it is clear the trip home takes 10 minutes less than when going to work. *careful about units* T is in hours. We must change 10 minutes to fractions of 1 hour. 10 min/60 min/hour = 1/6 hour the time to get back home is T - 1/6
phi
  • phi
average speed on the trip home is 8 miles faster. so going home you go (S+8) miles per hour that means your second equation is (S+8) (T-1/6) = 56

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phi
  • phi
average speed on the trip home? is going to be S+8 (speed going home) you have two equations and two unknowns so you can solve for S (and T) and then S+8
anonymous
  • anonymous
Thanks so much!!

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