Thus, $a =\frac{ dv }{ dt }=\frac{ dv }{ dx } \cdot \frac{ dx }{ dt }$ by chain rule

$\text { or, } a=v \frac{d v}{d x}=\left(3 \sqrt{12-x^2}\right) \cdot(-6 x) \cdot 1 / 2 \cdot\left(12-x^2\right)^{-1 / 2}$

At $x =3, a =3 \cdot \sqrt{3} \cdot(-18) / 2 \cdot \frac{1}{\sqrt{3}}=-27 m / s ^2$

Thus magnitude $=27$, hence $B$ is correct.

$A$ is wrong because acceleration is not uniform, but dependent on $x$. D is incorrect because maximum displacement means $v =0$, which gives $O =$ $12-x^2$ or $x=\sqrt{12}$