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So confused by this!

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle's position is given by x = 3.00 - 12.00t + 3t2, in which x is in meters and t is in seconds. Where is the particle when it momentarily stops?




    2. Relevant equations



    3. The attempt at a solution
    I tried plugging in m and s, like m= 3 - 12s +3s^2 and tried to solve for s but hit a dead end
     
  2. jcsd
  3. Sep 19, 2011 #2
    I know I'm probably doing it dead-wrong, though.
     
  4. Sep 19, 2011 #3

    SteamKing

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    What is the velocity of the particle when it stops?
     
  5. Sep 19, 2011 #4

    SteamKing

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    To reply to your attempted solution, why would you substitute 'm' for x and 's' for t?
    'm' and 's' are the units of the time variable t and the position variable x.
     
  6. Sep 19, 2011 #5
    I don't want to answer but can you give me a hint as to how I would find that?
    Well, if it's stopped...0?
     
  7. Sep 19, 2011 #6
    x = 3.00 - 12.00t + 3t2
    v= 12 +6t?
     
  8. Sep 19, 2011 #7
  9. Sep 19, 2011 #8
    ok so you have a function that describes the velocity (v = 12 + 6t) and you have a value for the velocity that you want to know what time it occurs at...
     
  10. Sep 19, 2011 #9
    So plug in 0 for v? 0=12 + 6t
    -2=t?
     
  11. Sep 19, 2011 #10
    When I then plug -2 into original? 39?
     
  12. Sep 19, 2011 #11
    HAHA, I see a mistake (when I took the derivative).
    v=-12 + 6t
    0=-12 + 6t
    2=t
     
  13. Sep 19, 2011 #12
    Now what?
     
  14. Sep 19, 2011 #13
  15. Sep 19, 2011 #14
    well done! :)
     
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