elevator going up with constant velocity but changing acceleration

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/s^sfor the first part

v = 9 m/s

and for the last part a = -1 m/s^s





2. Relevant equations



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



d=v_it + 1/2at²



and



v_f = v_i + 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=v_it + 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