Problema Solution
an elevator went from bottom to the top of a building at an average speed of four feet/second, remained at the top of the building for 90 seconds, and then returned to the bottom at five feet/second. If the total elapsed time was four and a half minutes, how high is the tower math help
Answer provided by our tutors
let
v1 = 4 ft/sec the average speed going up
v2 = 5 ft/sec the average speed going down
t1 = the time going up
t2 = the time going down
t = 4.5 min= 4.5*60 = 270 sec the total time
h = the height of the tower
the total time is:
t1 + 90 + t2 = 270
t1 + t2 = 270 - 90
t1 + t2 = 180
since the average speed = distance/time => distance = avg.speed*time (in our case the distance is the height of the tower)
t1*v1 = t2*v2
4t1 = 5t2
by solving the system of equations
t1 + t2 = 180
4t1 = 5t2
we find
t1 = 100 sec
t2 = 80 sec
click here to see the step by step solution of the system of equations
h = v1*t1 = v2*t2
h = 4*100
h = 400 ft
the tower is 400 feet high.