I'm doing this question for physics, and I'm a bit stuck.
The question is:
A hotel elevator ascends 200m with maximum speed of 5 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s^2
a) How far does the elevator move while accelerating to full speed from rest?
b) How long does it take to make the complete trip from bottom to top?
Part a I did
Vf^2 = Vi^2 +2aΔx
25 = 0 + 2(1)Δx
25/2 = Δx = 12.5 m (and I got the answer right)
now part b
I tried a couple of ways, I tried separating the problem after 12.5 m, so
x(t) = (.5 a t^2) + Vit + Xi
200 = .5t^2 + 5t + 12.5
So quadratic is .5t^2 +5t -187.5 = 0
t = 15 s
I added 5 s on top of that, since that is the time to the 12.5 m, from
Vf = Vi + a t
5 = 0 + 1(t)
but 20 s is apparently the wrong answer.
I tried treating the whole problem as one part, like
200 = .5at^2 + Vit + Xi
200 = .5t^2 + 0 + 0
400 = t^2
t=20 s
what am I doing wrong?
