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Velocity from position in a parametric

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img21.imageshack.us/img21/7910/68588225.th.jpg [Broken]


    2. Relevant equations



    3. The attempt at a solution

    I know that the velocity is the derivative of the position vector.. but I am kind of confused how to do this
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 29, 2009 #2
    How about graphing the table of values to start?
     
  4. Mar 29, 2009 #3
    I've done that as well... but from the graph I can't find the velocity
     
  5. Mar 29, 2009 #4
    You should have drawn two graphs (why?) and the velocity is the slope (why?).
     
  6. Mar 29, 2009 #5
    the velocity is the slope (why?).

    because it's the derivative of the graph...

    why two graph, I don't know
     
  7. Mar 29, 2009 #6
    You need 2 because you have 2 position components: x vs t and y vs t.
     
  8. Mar 29, 2009 #7
    ok, I got v = -4/5i + 5/5j, is this correct?

    if that's correct, how do I answer this one:

    Any time when the particle is moving parallel to the y-axis.
     
  9. Mar 29, 2009 #8
    I feel too lazy to verify that at moment. Just double-check the slope from each graph.

    This is where the x vs y graph comes into play. When would the particle by moving parallel to the y-axis in this graph? Hint: tangents.
     
  10. Mar 29, 2009 #9
    the graph looks really weird... it's like a circle
     
  11. Mar 29, 2009 #10
    Good. If it looks like a circle, there is a point where the tangent at that point is parallel to the y-axis. Do you agree?
     
  12. Mar 29, 2009 #11
    yes and that is at point x = 7 and y = 5.. which is at 7.5 sec

    how about:

    Any time when the particle has come to a stop?
     
  13. Mar 29, 2009 #12
    Hint: The particle is not moving when its velocity is 0.
     
  14. Mar 29, 2009 #13
    I got it now.. thanks for the help :)
     
    Last edited: Mar 29, 2009
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