Solving 2.52 Elevator Problem how to calculate time of a complete trip?

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SUMMARY

The 2.52 Elevator Problem involves calculating the total time for an elevator to ascend 200 meters with a maximum speed of 5 m/s, and both acceleration and deceleration at 1.0 m/s². The solution requires breaking down the trip into three phases: accelerating for 12.5 meters, cruising for 175 meters, and decelerating for the final 12.5 meters. Each phase's time can be calculated using kinematic equations, specifically xf = xi + vi Δt + 1/2aΔt² and v = vi + a*t. The total time is the sum of the time taken for each phase.

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crystaldreams
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Homework Statement


Hi everyone, I am just wondering how you would solve this question..
2. Elevator Problem (Practice on Segmented Motion)
A hotel elevator ascends 200 m with maximum speed of 5 m/s. Its acceleration and
deceleration both have a magnitude of 1.0 m/s
How long does it take to make a complete trip from bottom to top.

Homework Equations


xf = xi + vi Δt + 1/2aΔt^2
v = vi + a*t

The Attempt at a Solution


There was another part to this question which asked me to solve for how far the elevator will move while accelerating to full speed from rest, and I got 12.5 m for that.
Do I plug in 12.5 m into this equation and solve for t? Thank you!
 
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Welcome to PF crystaldreams!

Assuming you did the first part right (EDIT: and you did), then the elevator will accelerate upward for the first 12.5 m, cruise at the max speed for 175 m, and then accelerate downward (meaning decelerate: i.e. reduce its upward speed) for the final 12.5 m, bringing it to a rest at the top. All you have to do is calculate the time required for each of these three phases of motion using the kinematics equations.
 

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