Particle motion described by a time dependent function x(t)

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The particle's position is given by the function x(t) = 2.00 t^3 - 5.00 t + 1.00 m. At t = 4.00 s, the calculated position is 109 m. To find velocity and acceleration, users are advised to differentiate the position function. Velocity is the first derivative of the position function, while acceleration is the second derivative. Clarification on these calculations can help resolve confusion in solving the problem.
elizabethR
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how would i do this...
A particle moving along the x-axis has its position described by the function x=( 2.00 t^{3}- 5.00 t+ 1.00 )\; {\rm m}, where t is in s. At t= 4.00 , what are the particle's position- i calculated that to be 109m , velocity, and acceleration?

i have tried for over an hour on this one problem. i am clueless. my physics book sucks!
 
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Show some work, so we can point out what's wrong.

Btw, for the position at t=4, you already have the x(t) (position function), for velocity and acceleration, look up the definitions.

v = \dot{x}

a = \dot{v} = \ddot{x}
 
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