1. An object moves along the `x` axis according to the equation `x = 3.00t^2 - 2.00t + 3.00`, where `x` is in meters and `t `is in seconds. Determine (a) the average speed between `t = 2.00 s` and `t = 3.00 s`, (b) the instantaneous speed at `t = 2.00 s` and at `t = 3.00 s`, (c) the average acceleration between `t = 2.00 s` and `t = 3.00 s`, and (d) the instantaneous acceleration at `t = 2.00 s` and `t = 3.00 s`. (e) At what time is the object at rest?


Hint
`x = x_0 + v_0t + 1/2at^2`,`v= (dx)/(dt) =v_0 + at`,`a=(dv)/dt`
Answer
(a) `overline(v) = 13.0 m//s` (b) `v_(2.00s) = 10.0 m//s` `v_(3.00s) = 16.0 m//s` (c) `overline(a) = 6.00 m//s^2` (d) `a_(2.00s) = 6.00 m//s^2` `a_(3.00s) = 6.00 m//s^2` (e) `t=0.333s`
Show Steps
(a) `x_(2.00) = 3.00(2.00)^2-2.00(2.00) +3.00 =11.0m` `x_(2.00) = 3.00(3.00)^2-2.00(3.00) +3.00 =24.0m` `overline(v) = (24.0 m - 11.0 m)/(3.00s-2.00s) =color(red)(13.0 m//s)` (b) `v = (dx)/(dt) = 6.00t -2.00` `v_(2.00) = 6.00(2.00) -2.00 = color(red)(10.0 m//s)` `v_(3.00) = 6.00(3.00) -2.00 = color(red)(16.0 m//s)` (c) `a = (dv)/(dt) = 6.00` `v_(2.00) = color(red)(6.00 m//s^2)` `v_(3.00) = color(red)(6.00 m//s^2)` (e) when object is at rest `v=0 m//s` we get: `6.00t - 2.00 = 0` `t = 2.00/6.00 = color(red)(0.333 s)`