
#1
Sep1211, 05:22 PM

P: 1

A hotel elevator ascends 230 m with a maximum speed of 5.8 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s2.
(a) How far does the elevator move while accelerating to full speed from rest? I got this part: x = v/2a = 5.8/2(1) = 5.8^{2}/2 = 16.82 m (b) How long does it take to make the complete trip from bottom to top? I broke the problem up into three sections: the elevator going up (1.0 m/s^{2}), the elevator at rest (0), and the elevator coming back down (1.0 m/s ^{2}). I know that the elevator goes 230 m up and then 230 m down, but I cannot figure out which kinematic equation to use or if I need to use calculus (I have limited calculus knowledge). 



#2
Sep1211, 08:41 PM

PF Gold
P: 1,054

Think about how an elevator operates as it goes up: it accelerates, travels at constant speed for a while, then decelerates. Same 3 phases on the down trip. Don't worry about the elevator at rest partassume it goes up, stops for less than an instant, then goes back down.



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