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Math Help - Acceleration/velocity/Displacement problem

  1. #1
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    Acceleration/velocity/Displacement problem

    The acceleration of an object is given by the function a(t) = -t/2 + 9/4. Also at time t=0, the velocity of the object is -2. Find the difference between the distance and the displacement traveled by the object to the nearest integer from t=0 to t=10.

    Okay so this is what i have so far...

    I took the integral of the acceleration to get the velocity and what I obtained was 9t/4 - (t^2)/4 + C
    then I got -2 = C

    I then integrated the velocity equation to get the position equation and I got -(t^3)/12 + (9t^2)/8 -2t + C.

    How do I find C for the position equation?
    Or did I go to far...

    I know I need to set an equation equal to zero to determine min and max to see where the function changes direction and then I set up an integral of the same function from 0 to one point then from that point to 10...

    can someone please help me?
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  2. #2
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    The acceleration of an object is given by the function a(t) = -t/2 + 9/4. Also at time t=0, the velocity of the object is -2. Find the difference between the distance and the displacement traveled by the object to the nearest integer from t=0 to t=10.
    displacement ... \Delta x = \int_0^{10} v(t) \, dt

    displacement is a straight-forward FTC calculation.


    however, for distance traveled ... d = \int_0^{10} |v(t)| \, dt


    v(t) = \frac{-t^2 + 9t - 8}{4} = -\frac{(t-8)(t-1)}{4}

    v(t) = 0 at t = 1 and t = 8

    \int_0^{10} |v(t)| dt = -\int_0^1 v(t) \, dt + \int_1^8 v(t) \, dt - \int_8^{10} v(t) \, dt
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