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Particle position function

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data

    When will the velocity be zero?
    t is in minutes
    s is in metres



    2. Relevant equations

    s = 4t^3 - 8t^2 + t - 50




    3. The attempt at a solution

    I came up with the velocity equation by taking the derivative of function s.
    I came up with the acceleration equation by taking the derivative of the velocity function. I don't know how to figure out(solve) when the velocity of the particle will be zero? Help please?
    The velocity at t minutes is given by:
    v(t) = 12t^2 - 16t + 1
    Acceleration is given by:
    a(t) = 24t - 16
     
    Last edited: Jul 17, 2011
  2. jcsd
  3. Jul 17, 2011 #2

    Doc Al

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    Staff: Mentor

    Looks good. So plug in the value for v and solve the equation for t.
     
  4. Jul 17, 2011 #3
    can i use quadratic formula, like this?

    t = -b + - sqrt b^2 - 4ac/2a

    where a = 12, b = -16, and c = 1 ?
     
  5. Jul 17, 2011 #4
    Exactly, that will give you the times at which v(t) = 0
     
  6. Jul 18, 2011 #5

    Doc Al

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    Staff: Mentor

    Yes!
     
  7. Jul 18, 2011 #6
    My answer came to be:

    t = 16 + - sqrt208/24

    but the answer in the back of book is 4 + - sqrt13/6

    I am not sure how they got that?
     
  8. Jul 18, 2011 #7
    Try simplifying your square root expression more that you already have, and you'll find you have the same answer.
     
  9. Jul 18, 2011 #8
    oh yes thank you i see 16*13 = 208
     
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