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