4. A particle moves along the x axis. Its position is given by the equation `x=2+3t -4t^2`,with `x` in meters and `t` in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at `t = 0`.
(a)`2.56 m` (b)`-3m/s`
Given: `x = 2 + 3t - 4t^2` (a) from given equation we get: `v = (dx)/(dt) = 3 - 8t` when `v = 0` it changes direction, so ` 0 = 3-8t` `=> t =3/8 s` `x = 2 + 3(3/8) -4(3/8)^2 = color(red)(2.56 m)` (b) when `t=0` `x = 2 m` `2 = 2 + 3t - 4t^2` `t(3-4t) = 0` `t= 0` or `t = 3/4 s` When`t = 3/4 s` it returns to the position is had at t = 0 `v = 3 - 8t` `=>v = 3-8(3/4) = color(red)(-3m//s)`