A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2.
How long does it take to make the complete trip from bottom to top?
(a) From rest (u=0) .. v² = 2as .. s = (5m/s)² / (2 x 1m/s²) = 25 / 2 .. .. ►d = 12.50 m
This is the distance to which elevator move while accelerating
to full speed from rest
(b) acceleration time (= deceleration time) from .. d = ½.at²
(u=0)
.. t = √(2d/a) .. t = √(2 * 12.5m/s / 1m/s²) .. .. t = 5.0s [x2 for
accel.+decel.times = 10.0s]
Constant vel. time = d'/v = [200m - (2 x 12.5m)] / 5.0m/s = 175/5 =
35.0 s
Total time = (35.0 + 10.0) s .. .. ►T = 45.0
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