SOLUTION: An elevator went from the bottom to the top of a tower at an average speed of 4m/s, remained at the top for 90 s, and then returned to the bottom at 5 m/s. If the total elapsed was

Question 231060: An elevator went from the bottom to the top of a tower at an average speed of 4m/s, remained at the top for 90 s, and then returned to the bottom at 5 m/s. If the total elapsed was 4.5 min, how high is the tower?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
An elevator went from the bottom to the top of a tower at an average speed of
4m/s, remained at the top for 90 s, and then returned to the bottom at 5 m/s.
If the total elapsed was 4.5 min, how high is the tower?
:
Change 4.5 min to 270 sec
Elevator travel time,(remains at top for 90 sec):
270 - 90 = 180 sec
:
Let t = time for elevator to go to the top
Then
(180-t) = time to return to bottom
:
Write a distance (height) equation: dist = speed * time
Dist up = dist down
4t = 5(180-t)
4t = 900 - 5t
4t + 5t = 900
9t = 900
t = 100 sec
;
Use this value to find the dist up:
4 * 100 = 400 meters high
;
Check solution using return equation
5(180 - 100) = 400 meter high
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