Question 215308: A family drove 1080 miles to their vacation lodge. Bercause of increased traffic denisty, their average speed on the return trip was decreased by 6 miles per hour and the trip took 2 1/2 hours longer. Determine their average speed on the way to the lodge.
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! Velocity = Distance / Time,,,,or T=d/v
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t1 = d/v1 = 1080/v1
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but t2= t1+2.5,,,,,,v2=v1-6
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t2 = d/t2,,,,,(t1+2.5) = 1080/(v1-6),,,,,subst
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{(1080/v1) +2.5} = 1080/(v1-6)
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{ (1080+2.5v1)/v1} = 1080/(v1-6),,,,,mult thru by (v1-6) * (v1)
.(1080+2.5v1)*(v1-6) = 1080 *(v1)
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FOIL
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1080v1 -6*1080 +2.5v1^2 -6*2.5v1 = 1080v1
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.6480 +2.5v1^2 -15v1 =0
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2.5 v1^2 -15v1 +6480 =0,,,,,,divide by 2.5
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v1^2 -6v1 +2592 =0,,,,factor
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(v1 - 54)(v1-48) = 0
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v1= 48, 54
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check,,,(48),,,t1= 1080/48 = 22.5
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t2= 1080/ (48-6)=25.71,,,,3.2 difference,,,,not valid
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(54) t1=1080/54 = 20
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t2 = 1080/(54-6) = 22.5,,,difference = 2.5,,,,ok
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Therefore v1=54 MPH
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