Problema Solution
On the first part of a 27.5 mile trip,the average speed was 48 miles per hour. Later the average speed was reduced to 35 miles per hour. If one spent five times as long on the first part as on the second part what was the total time of the trip!
Answer provided by our tutors
let
d1 = 27.5 mi the distance of the first part of the trip
t1 = the time of the travel of the first part of the trip
v1 = 48 mph the average speed on the first part of the trip
v2 = 35 mph the average speed on the second part of the trip
t2 = 5t1 the time of the travel of the second part of the trip (one spent five times as long on the first part as on the second part)
we know that average speed = distance/time follows time = distance/avg.speed
t1 = d1/v1
t1 = 27.5/48 hr
we need to find the total time of the trip, that is, t1 + t2
t1 + t2 = t1 + 5t1 = 6t1 = 6*(27.5/48) = 3.4375 hr = 3 hr 0.4375*60 min = 3 hr 26.25 min = 3 hr 26 min 0.25*60 sec = 3 hr 26 min 15 sec
the total time of the trip was 3.4375 hour or 3 hr 26 min 15 sec.