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Intermediate Dynamics - HELP!

  1. Sep 7, 2010 #1
    1. A
     
    Last edited: Sep 7, 2010
  2. jcsd
  3. Sep 7, 2010 #2

    Doc Al

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    Find the acceleration, then use a bit of kinematics.
     
  4. Sep 7, 2010 #3

    Doc Al

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    Since you don't have numbers, you'll express your answer in terms of the given data--F0, t0, and m.

    You're on the right track. Hint: Treat each segment of the motion separately. In the first segment (from t = 0 to t0), the initial speed is zero--but that's not the case for the second half of the motion.

    Hint2: Consider the average speed during each segment.
     
  5. Sep 7, 2010 #4

    Doc Al

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    There's a much easier way to find the final velocity of the first segment. What's the definition of acceleration?
     
  6. Sep 7, 2010 #5
    For a final answer I got

    x=((2F0*t0^2)/m)) + (4F0t0)/m

    But... I don't think thats right at all...
     
  7. Sep 7, 2010 #6

    Doc Al

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    No, it's not.

    Do it step by step, one segment at a time.
     
  8. Sep 7, 2010 #7
    Ok what I did was :

    First I used x=v0t+1/2at^2 to find the distance traveled for the first segment before the force doubled

    I plugged in a=(F/m) for the acceleration and v0=0 for the initial velocity
    Then i got
    x=1/2(F0/m)t^2
    for the first segment

    Then,
    i used vf^2=vi^2+2ax to find the final velocity of the first segment
    (plugging in the x i found from above and a=F/m and v0=0)
    which i got to be
    vf= (F0t)/m


    then I used that final velocity as the initial velocity for the 2nd segment
    and plugged in a=2F0/m
    into x=v0t+1/2at^2


    and somehow came out with the answer
    x=((2F0*t0^2)/m)) + (4F0t0)/m


    Where did I go wrong ?
     
  9. Sep 7, 2010 #8
    The definition of acceleration is the change in velocity over the change in time
     
  10. Sep 7, 2010 #9

    Doc Al

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    Good! You have the first segment done.

    Good. An easier way would be to use vf = vi + at, but your way is perfectly fine.


    That will give you the distance for the second segment. (Then you'll add that to what you already found for the first segment.)


    I don't see how you got this. Show me what you plugged in where.
     
  11. Sep 7, 2010 #10
    Ooh, well I never went back and added the first segment.. maybe thats where I went wrong ? .. and my algebra could be a little off too. I'll probably need to go back and double check that also.
     
  12. Sep 7, 2010 #11

    Doc Al

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    Redo and then simplify your results for the second segment.
     
  13. Sep 7, 2010 #12
    For the second segment I'm getting the distance as

    [F0(3t0 + t0^2)]/m
     
  14. Sep 7, 2010 #13

    Doc Al

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    That's not dimensionally correct. Show me what you plugged in for each term in the following:
    x=v0t+1/2at^2
    x = (v0)(t) + 1/2(a)(t)^2
     
  15. Sep 7, 2010 #14
    v0 ---> (F0t0)/m
    t ----> 2
    a -----> (2F0)/m
     
  16. Sep 7, 2010 #15

    Doc Al

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    v0 and a are correct, but the time for the second segment is the same as the first: t0. (From t0 to 2t0.)
     
  17. Sep 7, 2010 #16
    Oooh. That makes sense. Duh.
     
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