1. A particle moves along the `x` axis according to the equation `x=2.00+3.00t -1.00t^2`, where `x` is in meters and `t` is in seconds. At `t = 3.00 s`, find (a) the position of the particle, (b) its velocity, and (c) its acceleration.
`x = x_0 + v_0t + 1/2at^2`,`v= (dx)/(dt) =v_0 + at`
(a) `2.00 m` (b) `-3.00m//s` (c) `-2.00 m//s^s`
Givens: `x=2.00+3.00t -1.00t^2` `t = 3.00s` (a) at `t=3.00s` `x = 2.00 + 3.00(3.00) - 1.00(3.00)^2 = color(red)(2.00m)` (b) `v = (dx)/(dt) = 3.00 -2.00t` at `t = 3.00s` `v = 3.00 - 2.00(3.00) = color(red)(-3.00 m//s)` (c) `a = (dv)/(dt) =-2.00m//s^2` at `t=3` `a = color(red)(-2.00m//s^2)`