Elevator going up with constant velocity but changing acceleration

1. Jan 29, 2014

alexs2jennisha

1. The problem statement, all variables and given/known data
An elevator goes up 180 m by first accelerating at a constant rate of 1.0 m/s2, then staying at a constant
speed of 9 m/s and then decelerating at a constant rate of -­‐‑1.0 m/s2. How much time does it take the
elevator to go from bottom to top?

so i know that d = 180m
a = 1.0 m/ssfor the first part
v = 9 m/s
and for the last part a = -1 m/ss

2. Relevant equations

Im not sure which kinematics to use, there are two that I tried using

d=vit + 1/2at2

and

vf = vi + at

3. The attempt at a solution

I tried to find the time for the part where the elevator is accelerating by doing

9 = 0 + 1t

so t = 9s

then I tried to find the distance for which the elevator was going up by doing

d = 0(9) + 1/2(1)(92) and got d = 40.5

so then the d for when its decelerating is 139.5 which i plugged into

d=vit + 1/2at2

and tried to solve for t but i ended up getting a weird answer with a square root.

am i approaching the problem the right way? or am i completely wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 29, 2014

ehild

correct so far...

It also travels some distance with constant velocity...

ehild