SOLUTION: 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 too

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
.
t1 = d/v1 = 1080/v1
.
but t2= t1+2.5,,,,,,v2=v1-6
.
t2 = d/t2,,,,,(t1+2.5) = 1080/(v1-6),,,,,subst
.
{(1080/v1) +2.5} = 1080/(v1-6)
.
{ (1080+2.5v1)/v1} = 1080/(v1-6),,,,,mult thru by (v1-6) * (v1)
.(1080+2.5v1)*(v1-6) = 1080 *(v1)
.
FOIL
.
1080v1 -6*1080 +2.5v1^2 -6*2.5v1 = 1080v1
.
.6480 +2.5v1^2 -15v1 =0
.
2.5 v1^2 -15v1 +6480 =0,,,,,,divide by 2.5
.
v1^2 -6v1 +2592 =0,,,,factor
.
(v1 - 54)(v1-48) = 0
.
v1= 48, 54
.
.
check,,,(48),,,t1= 1080/48 = 22.5
.
t2= 1080/ (48-6)=25.71,,,,3.2 difference,,,,not valid
.
(54) t1=1080/54 = 20
.
t2 = 1080/(54-6) = 22.5,,,difference = 2.5,,,,ok
.
Therefore v1=54 MPH
.

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