Problema Solution
A messenger takes a freight elevator up to the 42nd floor and returns in a passenger elevator .
the freight elevator travels a speed of 1 foot in 3 seconds; the passenger elevator,
at 3 feet per second. Find the average speed for the entire trip in feet per minute
Answer provided by our tutors
the average speed over a finite time interval is the total distance traveled divided by the time duration
avg speed = total distance / total time
let
d = the distance in feet
v1 = 1 ft / 3 sec = 1ft / 3*60^(-1) min = 20 ft/min since 1 sec = 60^(-1) min the speed of the freight elevator
v2 = 3 ft / sec = 3ft / 60^(-1) min = 180 ft/min since 1 sec = 60^(-1) min the speed of the passenger elevator
time going up = d/20 min
time going down = d/180 min
avg speed = 2*d / (d/20 + d/180) = 2 / (9/180 + 1/180) = 2*180/10 = 36 ft/min
the average speed for the entire trip was 36 ft/min.