Problema Solution
Mr. Johnson made a certain trip in two hours, while returning it took half an hour longer. If his average speed going was 6 miles per hour more than than his average returning, what was his average speed going and his average speed returning
Answer provided by our tutors
let
v1 = the average speed going
t1 = 2 hr the time of the trip going
v2 = the average speed returning
t2 = 2 + 1/2 = 2.5 hr the time of the trip returning
his average speed going was 6 miles per hour more than than his average returning
v1 = 6 + v2
since speed = distance/time => distance = speed*time
v1*t1 = v2*t2
2v1 = 2.5v2
by solving the system of equations
v1 = 6 + v2
2v1 = 2.5v2
we find
v1 = 30 mph
v2 = 24 mph
click here to see the step by step solution of the system of equations
the average speed going is 30 mph.
the average speed returning is 24 mph.