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Position and Velocity as a function of time

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data
    At time t=0 an ant starts to walk from -2f to -f at a constant speed v. Show that the position (q) and velocity (u) of the ants image as a function of time is given by:

    q=(2f-vt)f/(f-vt) AND u = f2v/(f-vt)2

    cBWUY.jpg
    2. Relevant equations

    d(u/v)/dx = [V(du/dx)-U(dv/dx)]/v2

    3. The attempt at a solution

    I am finding it hard to start the solution.
     
    Last edited: Nov 18, 2012
  2. jcsd
  3. Nov 18, 2012 #2

    haruspex

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    When the ant is at distance x from the lens, where is the image? You surely have a standard equation for that.
     
  4. Nov 18, 2012 #3
    1/q=1/f+1/x
    where x is the ants distance
    q is the image distance and
    f is the focal length

    OR In newtonian terms:
    X1 = S1 - f
    X2 = S2 - f
     
    Last edited: Nov 18, 2012
  5. Nov 18, 2012 #4

    haruspex

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    OK, so if you can express the ant's position as a function of time, you can use the above equation to get the position of the image as a function of time.
     
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