physics

The position of a particle moving along the x axis is given by x=3.0t^2 - 1.0t^3 , where x is in meters and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction

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  1. If x is the displacement, note that
    v (velocity) = dx/dt, and
    a (acceleration) = dv/dt
    Therefore, we get the derivative of the given equation:
    dx/dt = 6t - 3t^2 = velocity
    Note that this is just the equation for velocity, and we need to maximize it. Thus, we differentiate it again:
    dv/dt = 6 - 6t = acceleration
    We equate this to zero and solve for t:
    0 = 6 - 6t
    6t = 6
    t = 1 s
    At t = 1, the velocity is maximum, and the displacement is
    x = 3.0t^2 - 1.0t^3
    x = 3*(1^2) - (1^3)
    x = 3 - 1
    x = 2 m

    Hope this helps :3

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