Problema Solution

Alberto made a trip to his friend's house and back. The trip there took three hours and the trip back took four hours. He averaged 9.8 mph faster on the trip there than on the return trip. Find Alberto's average speed on the outbound trip.

Answer provided by our tutors

let


v = the speed on the return trip

v + 9.8 = the speed on the trip there

t1 = 3 hr the time of the trip there

t2 = 4 hr the time of the return trip



since speed = distance/time => distance = speed*time


v*t2 = (v + 9.8)*t1


4v = (v + 9.8)*3


by solving we find


v = 29.4 mph


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the speed of the outbound trip is 29.4 + 9.8 = 39.2 mph.