Answer
$a=-9.00cm/s^{2}$
Work Step by Step
Take the second derivative of the position function, or the derivative of the velocity function, to determine the acceleration as a function of time:
$\frac{d^{2}}{dt^{2}}x(t)=\frac{d^{2}}{dt^{2}}(9.00t-0.750t^{3})$
$\frac{d}{dx}v(t)= \frac{d}{dt}(9.00-2.25t^{2})$
$a(t)=-4.50t$
Plug in the time when the particle momentarily stops $(t=2s)$:
$a(2)=-4.50(2)$
$a(2)=-9.00cm/s^{2}$
