Elevator going up with constant velocity but changing acceleration

[alexs2jennisha]

Homework Statement

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

Homework 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

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?

 

[ehild]

Homework Statement

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

Homework 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

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

correct so far...

so then the d for when its decelerating is 139.5

It also travels some distance with constant velocity...


ehild