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Please check my work on this short calc (?) prob

  1. Sep 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Two particles move along an x axis. Position of particle 1 is x = 6t^2 + 3t + 2. The acceleration of particle 2 is given by a = -8t and at t=0, its velocity is 20 m/s. When the velocities of the particles match, what is their velocity?


    2. Relevant equations



    3. The attempt at a solution

    Ok, so I took the derivative of the position function and found it to be v = 12t + 3. Then I took the integral from the acceleration function and found it to be v = -4t^2 + C ( by plugging t=0, I got C = 20). Hence, v = -4t^2 + 20.

    Now, I equated both velocity functions and was left with 4t^2 + 12t -17. Does that mean I should solve the quadratic to get two answers--this seems too simple and the problem is supposed to be 3/3 for difficulty...
     
  2. jcsd
  3. Sep 4, 2007 #2

    yes, that's what I did also, before reading yours work.
     
  4. Sep 4, 2007 #3

    Doc Al

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    Staff: Mentor

    Your work looks right to me.
     
  5. Sep 4, 2007 #4
    Thank you, but how is it that you come out with two aswers? Should I eliminate the negative one or is it physically possible in this case?
     
  6. Sep 4, 2007 #5
    because it's quadratic v = -4t^2 + 20

    why you hate negative value lol?
    how's it different from positive?
     
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