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Relative velocity of bolt

  1. Mar 3, 2013 #1
    1. The problem statement, all variables and given/known data

    An elevator is moving upward at a constant speed of 2.50 m/s. A bolt in the elevator ceiling 3.00m above the elevator floor works loose and falls.

    A. How long will it take the bolt to fall to the elevator floor?

    I am not sure if I should do this relative to elevator or earth.

    2. Relevant equations

    Vb/g=Vb/e+Ve/g

    Yb/g=Yb/e+Ye/g

    3.The attempt at a solution
    Vb/g=Vb/e+Ve/g
    Vb/g= velocity of bolt relative to ground
    Vb/e= velocity of bolt relative to earth
    Ve/g= velocity of elevator relative to ground


    Yb/g=Yb/e+Ye/g
    Yb/g=position of bolt relative to ground
    Yb/e= position of bolt relative to earth
    Ye/g=position of elevator relative to ground


    Vb/g=Vb/e+Ve/g
    Vb/g=0+2.5m/s

    Yb/g=Yb/e+Ye/g
    Yb/g=3m+ 2.5m/s*t

    Is this right so far? Where do I go from here? Thanks!
     
  2. jcsd
  3. Mar 3, 2013 #2

    haruspex

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    You can make this question much easier by recognising that since the elevator is moving at constant speed it constitutes an inertial frame.
     
  4. Mar 3, 2013 #3
    Can you clarify what you mean? :/
     
  5. Mar 3, 2013 #4

    haruspex

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    Take the elevator as the reference frame. What are the initial height, initial velocity, final height and acceleration of the bolt in that frame?
     
  6. Mar 3, 2013 #5
    3m, 0m/s, 0m, -9.81m/s^(2)?

    I think I see what you mean. What if the elevator was not traveling at a constant speed but rather a constant acceleration. How would that change?
     
  7. Mar 3, 2013 #6

    tms

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    He means that the laws of physics are the same in every inertial reference frame. You can think of an inertial frame as a frame that is not accelerating (that is, one that is moving at constant speed relative to another inertial frame). It's a little more complicated than that, but you can ignore those complications for present purposes.
     
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