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Acceleration of an elevator

  1. Nov 18, 2007 #1
    I need help getting started with this problem. Thanks.

    1. The problem statement, all variables and given/known data

    The acceleration of an elevator changes at a rate give by the function


    Plot the position of the elevator assuming that its velocity at t=0 is zero and the elevator starts moving from a zero ground level.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 18, 2007 #2


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    Are you in a calculus-based physics course? The problem sounds like it calls on you to integrate this acceleration function to find the velocity function, then to integrate that to find a positive function, given that v(0) = 0 and x(0) = 0.

    This will be harder to deal with if you have to do it only graphically.
  4. Nov 18, 2007 #3
    I could use integration to solve this, then I have to use the values to plot a graph.
    Still need help staring this problem. Thanks.
  5. Nov 18, 2007 #4


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    So wouldn't you be plotting the values of x(t) for different times? Also, is e in that function supposed to be 2.71828... ? So you have all the information you need to find the position function and calculate positions, no?
  6. Nov 22, 2007 #5
    e = 2.71828183

    I have integrated the acceleration equation twice to find out position. Then I plotted the equation on Maple and i got this graph, Can someone tell weather this is correct or incorrect way of addressing this problem. I also can't understand the Graph, according to the graph it says the the position of the elevator increases exponential after 5 second. Can someone explain the graph to me. Thanks

    Last edited: Nov 22, 2007
  7. Nov 22, 2007 #6
    - First of all it does not make sense to plot the position time graph for negative time values. You should graph the position function for t>=0

    - Secondly as for explaining the graph for positive t values note the following:
    The elevator at t = 0 starts at about 14 units away from the origin of your frame of reference and for t = 4 it will come towards the origin and then for t>4 it will move away from the origin.
  8. Nov 22, 2007 #7
    Thanks for the reply

    -The reason i plotted the graph for negative values is to get a perspective of what is happening.

    -Is the "y axis" refering to the floor number?

    -Also why is it moving away from the orgin at a expoential rate?
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