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Particle Movement

  1. Oct 9, 2009 #1
    1. A particle moves in such a way that its position z, in meters, is given as a function of t, in seconds, by the equation z = 2.12m/s^2t^2 - 2.96m/s^3t^3

    What is the velocity (m/s) of the particle at t = 0.657s

    The answer is -1.05, which is not what I got.

    2. First, I used the equation Xf = Xi + 1/2 (Vi + Vf) (t)

    Plugging in .0757 for Xf

    My answer was V=.231

    Then I tried V= Displacement/time

    .0757/.657 = .115

    What am I doing wrong here?
     
  2. jcsd
  3. Oct 9, 2009 #2

    kuruman

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    You are using the wrong equation.

    Xf = Xi + 1/2 (Vi + Vf) (t) is valid only when the acceleration is constant. Here it is not. How do you find the velocity as a function of time if you know the position as a function of time? Hint: Velocity is the rate of change of position with respect to time.
     
    Last edited: Oct 9, 2009
  4. Oct 9, 2009 #3
    I already tried the equation

    V=Displacement/time

    .0757/.657 = .115

    Do I now have to substitute that answer in to the equation for position as a function of time?
     
  5. Oct 9, 2009 #4
    or...do I have to take the derivative of v=displacement/time
     
  6. Oct 9, 2009 #5

    kuruman

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    Yes, you have to take the derivative. v = dz/dt.
     
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