The discussion focuses on calculating the total time for an elevator to travel a distance of 199 meters, starting and stopping at rest, with a maximum speed of 313 m/min and an acceleration of 1.10 m/s². Participants outline the three stages of motion: accelerating to maximum speed, traveling at constant speed, and decelerating back to rest. Key calculations include the distance covered during acceleration (12.3698 m) and the remaining distance (174.2604 m) to be traveled at maximum speed. The final time calculated for the trip was 919.49 seconds, prompting further inquiry into the accuracy of the numbers used.
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Doc Al said:What have you done so far? How quickly does it reach maximum speed? What distance does that require?
Good.mossfan563 said:Well part A was how far does the cab move while accelerating to full speed from rest?
And the answer was 12.3698 m.
That's what I've done so far.
Doc Al said:Good.
Treat the motion as having three stages: Accelerating to max speed; constant speed; decelerating to zero speed.
Figure out the distance and time associated with each stage of the motion.
mossfan563 said:Well I technically have the distance and time for accelerating to max speed. I can't figure out the other two though. I tried the quadratic equation and it doesn't work.
LowlyPion said:What quadratic equation? You have 3 regions. There are the two - one at each end - getting to max speed and going back to 0, and you know the distances for both because they are symmetrically the same.
Since you know the distance for both, then you know what remains of the trip that you will account for with max speed. Knowing the distance and knowing the max speed then you can figure the time.
Time for the trip? Total time = Time to speed + time at max + time to slow to 0.
mossfan563 said:Ok I understand that the time to speed and time to slow to 0 are pretty much the same. The time at max still is still confusing me. The remaining distance would be around 174.2.
I do know the max speed. I just don't know what formula to use.
mossfan563 said:Ok I used this formula.
Here are the numbers I used.
Total distance = 199 m
Remaining distance = 174.2604 m
Time of speed up and slow down = 4.742... s
Max speed = 5.21666666... m/s
Using all these numbers I got an answer of 919.49 s.
Is this off? What am I doing wrong? Are my numbers wrong?