# 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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