On a recent trip, sarah'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.

Let x mph be the speed for the 110 miles trip. Then for 80 miles,
Speed = (x-20) mph

Time, t = Distance/Speed
On 110 miles, t1 = 110/x hrs
On 80 miles, t2 = 80/(x-20) hrs

Total time = 4 hrs = t1+t2 = 110/x + 80/(x-20)

Solving for x;
4 = [110(x-20) + 80(x)]/x(x-20)
4(x)(x-20) = 110x -2200 + 80x
4x^2 - 80x = 110x - 2200 + 80x
4x^2 -80x -110x  - 80x +2200 = 0
4x^2 -270x +2200 = 0
Solving the quadratic equation;
x = [-(-270)+/- Sqrt ((-270)^2 -4(4)(2200)]/2*4 = 33.75+/- 24.27 = 9.48 mph or 58.02 mph

Ignore the value smaller than 20 as this would yield a negative value of second speed which is not practical.

Therefore, speed in the first 110 miles is 58.02 mph

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