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Velocity and distance traveled of a particle

  1. Feb 10, 2009 #1


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    1. The problem statement, all variables and given/known data
    (I can't seem to get the super and subscripts to work so I apologize if this looks ugly.) The position of a particle moving along the x-axis is given at time t by: x = C0 + C1t + C2t^2. At time t = 4.0s the acceleration of the object is a = -20 m/s^2 and that the object passes through the origin x = 0 at t = -2.0s and t = 6.0s. Find the velocity of the particle at t = -2.0s and the maximum value of the distance of the particle from the origin measured in the positive x-direction.

    2. Relevant equations
    x = (0.5)at^2 + v0t + x0

    3. The attempt at a solution
    I've tried three approaches: differentiating with respect to the x equation above, integrating from the acceleration given, and setting up x = 0 = (t + 2)(t - 6) from the information given.

    Differentiating and integrating lead me to the same problem.. how to find the constant C in v(t) = -20t + C.

    Setting up the factor-like problem I obtain t^2 -4t - 12. Looking at this equation I'm led to believe that C2 = 1, C1 = -4, and C0 = -12 from the coefficients. Taking the derivative of this, I get v(t) = 2t - 4. Deriving again I get 2 ms^2 which is not the -20 ms^2 provided.

    I'm not sure how to start solving this equation correctly. Could someone point me in the right direction? Thanks.
  2. jcsd
  3. Feb 10, 2009 #2


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    Gold Member

    I am a bit too lazy to completely work your problem out but from the information given, the acceleration is constant. So you simply differentiate twice to find that [tex]2C_{2} = -20 [/tex]

    edit: just made it a bit prettier
  4. Feb 10, 2009 #3


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    Yes, C2 = -10 in my calculation as well and substituting -10 back into v(t) = 2(C2)t + C1, I get v(t) = -20t + C1. How do I find C1 so that I can substitute -2 for t to solve for v(t)? Or is my answer for the velocity of the particle at time t = -2s simply v(-2) = 40 + C1 m/s? If I don't have a value for C1 then how do I find the maximum value of the distance of the particle in my original x equation?
  5. Feb 10, 2009 #4


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    Homework Helper

    You know where it is at t = -2 and at t = 6

    It must have reversed direction in that 8 seconds. That means at T = 2 it's |V| is 0.

    Since a is constant then |v| = a*t = a*4sec at the origin.
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