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)`
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