SOLUTION: Solve the problem. On a recent trip, Sarah's car traveled 20 mph faster on the first 140 miles than it did on the remaining 80 miles. The total time for the trip was 4 hours. Find

Question 371019: Solve the problem. On a recent trip, Sarah's car traveled 20 mph faster on the first 140 miles than it did on the remaining 80 miles. The total time for the trip was 4 hours. Find the speed of Sarah's car on the first part of the trip.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let speed on second leg be x mph
speed on first leg = x+20
..
second leg distance = 80 miles
first leg distance = 140 miles
..
time first leg + time second leg = 4 hours
...
t= d/t
80/x + 140/(x+20)=4
(80(x+20)+140x)/x(x+20)=4
80x+1600+140x=4x(x+20)
220x +1600 = 4x^2+80x
4x^2+80x-220x-1600=0
4x^2-140x-1600=0
/4
x^2-35x-400=0
solve for x using quadratic formula
discriminant = 2825
x1= (35+sqrt(2825))/2
x2 will be negative so to be ignored
= 44.07 mph on the second leg.
first leg speed = x+20 = 64.07 mph
..
m.ananth@hotmail.ca
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