How to find the maximum velocity and maximum acceleration?

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To find the maximum positive coordinate, velocity, and acceleration of a particle described by the position function x=12t^2-2t^3, the first derivative for velocity, V=24t-6t^2, is set to zero to determine critical points. Substituting these critical points back into the position function yields the maximum positive coordinate. For maximum velocity, the same process is applied to the velocity function, while the maximum acceleration is found by taking the derivative of the acceleration function and setting it to zero. The fourth derivative of position, known as jerk, is also mentioned as part of the discussion. Understanding these derivatives is crucial for solving the problem effectively.
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Homework Statement


The position of a particle moving along an x-axis is given by x=12t^2-2t^3, what is the maximum positive coordinate, velocity and acceleration reached by the particle?

Homework Equations





The Attempt at a Solution


I took derivative of the quadratic and got V=24t-6t^2 and set it =0, then i solve for t and substituted in x=12t^2-2t^3 to get the max. positive coordinate.
But I don't know what to do to solve for the max velocity and acceleration:confused:
 
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You have the velocity as a function of time, just like you had the position as a function of time.
 
more hints?
 
Why not repeat the same process for your function of velocity?
 
Oh, so i set a=0 to find max velocity, but then what do I do to find the max acceleration?
 
Find the derivative of acceleration as a function of time and set it to zero. The physical term for the fourth derivative time is called jerk.
 

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