SOLUTION: on a recent trip, Sararh's car traveled 20 mph faster on the first 110 miles than it did on the remaining 80 miles. The total time for the trip was 4 hr. Find the speed of Sarah's

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Question 336523: on a recent trip, Sararh's car traveled 20 mph faster on the first 110 miles than it did on the remaining 80 miles. The total time for the trip was 4 hr. Find the speed of Sarah's car on the first part of the trip.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Rate*Time=Distance
.
.
.
R1%2AT1=D1
R2%2AT2=D2
.
.
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D1=110%0D%0A%7B%7B%7BD2=80
T1%2BT2=4
R1=R2%2B20
.
.
.
Substituting,
%28R2%2B20%29%2AT1=110
T1=110%2F%28R2%2B20%29
.
.
R2%2AT2=80
T2=80%2FR2
.
.
And finally,
T1%2BT2=4
110%2F%28R2%2B20%29%2B80%2FR2=4
110%2B80%28R2%2B20%29=4%2AR2%2A%28R2%2B20%29
110R2%2B80R2%2B1600=4R2%5E2%2B80R2
4R%5E2-110R2-1600=0
Use the quadratic formula,
R2+=+%28110+%2B-+sqrt%28+%28110%29%5E2-4%2A4%2A%28-1600%29+%29%29%2F%282%2A4%29+
R2+=+%28110+%2B-+sqrt%2812100%2B25600+%29%29%2F%288%29+
R2+=+%28110+%2B-+sqrt%2837700%29%29%2F%288%29+
R2+=+%28110+%2B-+10sqrt%28377%29%29%2F%288%29+
Only the positive result makes sense for this problem.
R2=%28110%2B10sqrt%28377%29%29%2F8
So then
R1=%28110%2B10sqrt%28377%29%29%2F8%2B20
R2=%28270%2B10sqrt%28377%29%29%2F8 or approximately,
highlight%28R2=58.02%29mph