Solve Elevator Motion/Time: 230m, 5.8m/s, 1.0m/s2

[Cman1494]

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) = 5.8²/2 = 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/s²), 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).

 

[lewando]

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 part--assume it goes up, stops for less than an instant, then goes back down.