#### Answer

(a) $t = 0$ or $t = 3~s$
(b) At t = 0:
$x = 12~m$
$a = -18~m/s^2$
At t = 3 s:
$x = -15~m$
$a = 18~m/s^2$

#### Work Step by Step

(a) $x(t) = 2t^3-9t^2+12$
$v(t) = \frac{dx}{dt} = 6t^2-18t$
We can find $t$ when $v = 0$
$6t^2-18t = 0$
$(6t - 18)~t = 0$
$t = 0$ or $t = 3~s$
(b) $x(t) = 2t^3-9t^2+12$
$a(t) = \frac{dv}{dt} = 12t-18$
At t = 0:
$x = 12~m$
$a = -18~m/s^2$
At t = 3 s:
$x = 2(3)^3-9(3)^2+12$
$x = -15~m$
$a = (12)(3)-18$
$a = 18~m/s^2$