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Position, Acceleration, and Velocity of an Elevator

  1. Aug 26, 2008 #1
    1. The problem statement, all variables and given/known data
    A hotel elevator ascends 200 m with a maximum speed of 4.8 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s2.

    (a) How far does the elevator move while accelerating to full speed from rest? (Answer in Meters)

    (b) How long does it take to make the complete trip from bottom to top? (Answer in seconds)

    2. Relevant equations
    Velocity= delta-x/ delta-t

    Acceleration= Delta-v/ delta-t


    3. The attempt at a solution

    For some reason I can't reason this one out. I have attempted to graph the velocity vs. time graph (which I believe will hold the solutions) and my graph turned out to be a triangle with a little below the x-axis. Yet this graph doesn't look quite right to me, and I can't figure out how to derive the correct velocity vs. time graph from the given information. I have also tried to draw the acceleration graph, slowly speeding up the elevator (1 m/s^2), but my graph has almost leveled out, proving something must be entirely wrong. Pure math isn't doing it for me. Any ideas?
     
  2. jcsd
  3. Aug 26, 2008 #2

    LowlyPion

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    Homework Helper

    Maybe break it up into several problems? How long does it take to accelerate to the Vmax. That's almost the first part of the problem you have to find anyway. Happily it takes the same time to slow down at the same rate.

    Once you found the time to speed up you can find the distance to speed up. That's the answer to part A.The distance to slow down is again happily the same. Then you know the distance that you will travel at Vmax and hence the time.

    The formulas here may also help you:
    https://www.physicsforums.com/showpost.php?p=905663&postcount=2
     
  4. Aug 26, 2008 #3
    That really seemed like a perfect way to attack the problem after I read your post, but I can't figure out how long it takes to accelerate. I mean, I think the wording is confusing me. Do they mean the is takes just 1 m/s^2 to accelerate to 4.8 m/s? And then, how can I translate m/s^2 into velocity? Is that even possible with just the number 1? I'm completely flabbergasted on this one. Any other suggestions?
     
  5. Aug 26, 2008 #4

    LowlyPion

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    Homework Helper

    The statement of the problem tells you the acceleration. Velocity from 0 is time * acceleration. Hence time is Velocity / acceleration. Time to max speed then it 4.8 / 1.

    Distance is related by the equation x = 1/2 a*t2

    These equations are at the link I provided.
     
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