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Instantaneous velocity and speed problem

  1. Aug 31, 2008 #1
    1. The problem statement, all variables and given/known data
    the position function x(t) of a particle moving along an x axis is x=4.0-6.0t^2, with x meters and t in seconds. (a) at what time and (b) where does the particle (momentarily)stop?


    x=4.0-6.0t^2

    i made x=0
    and i made the equation equal this t=sqrt(4.0/6 = .8164995809
    i put it into the equation and x=0

    is this correct. I'm assuming that the particle is zero meters when it stops
    but what do i know i'm not an expert
     
  2. jcsd
  3. Aug 31, 2008 #2

    Doc Al

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    Staff: Mentor

    You can't just assume the answer. Instead, figure it out. Hint: Given the position as a function of time, find the velocity as a function of time.
     
  4. Aug 31, 2008 #3
    So are you saying that I have to find the velocity as a function of time.
    I'm not sure by what you mean by this. Do you mean that I use V=change inx/ change in t
     
  5. Aug 31, 2008 #4
    Are you saying that I have to find the derivative.
     
  6. Aug 31, 2008 #5

    Doc Al

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    That would be most wise. :wink:
     
  7. Aug 31, 2008 #6
    Ok, so would it be like this x=4.0-6.0t^2
    x=-(2)(6)(t)
    x=-12(t)



    so x=-12t

    then t=x/-12
     
    Last edited: Aug 31, 2008
  8. Aug 31, 2008 #7
    then i plug 12 into the equation and i get 4.0-6(12)^2=-860m
     
  9. Aug 31, 2008 #8

    Doc Al

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    That gives you the velocity as a function of t (not x). v = dx/dt.
     
  10. Aug 31, 2008 #9
    So how do I do that.
     
  11. Aug 31, 2008 #10
    hmm. would it be -12=x/t
     
  12. Aug 31, 2008 #11

    Doc Al

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    Do what? You already found the velocity, you just mislabeled it "x" instead of "v".
     
  13. Aug 31, 2008 #12
    oh ok lol, so that's all i do, v=-12t
     
  14. Aug 31, 2008 #13

    Doc Al

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    Good. Now go back and answer part (a).
     
  15. Aug 31, 2008 #14
    Ok so where do I plug in the -12t,
    x=4.0-6.0t^2

    would it be 4-6(-12)^2=-860
     
  16. Aug 31, 2008 #15

    Doc Al

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    You don't plug it in anywhere, you use that equation for speed to answer question (a). What's the time when the particle momentarily stops? (What speed must it have when it stops?)
     
  17. Aug 31, 2008 #16
    the speed would have to be zero right.
     
  18. Aug 31, 2008 #17
    -12(0)= it has to be zero
     
  19. Aug 31, 2008 #18

    Doc Al

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    Of course. What value of t gives you a speed of zero?
     
  20. Aug 31, 2008 #19
  21. Aug 31, 2008 #20
    That means that the particle stops at 4 meters right.
     
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