Particle Acceleration in Constant Potential Region

The force $F(x)$ acting on a particle moving along the $x$-axis in a conservative force field is related to its potential energy function $V(x)$ by the relation:
$F(x) = -\frac{dV(x)}{dx}$

According to the given potential energy profile, for $|x| > X_0$, the potential energy becomes constant:
$V(x) = V_0$

Differentiating $V(x)$ with respect to position $x$ in this region:
$\frac{dV(x)}{dx} = \frac{d}{dx}(V_0) = 0$

Thus, the net force acting on the particle is:
$F(x) = 0$

By Newton's second law of motion, the acceleration $a$ of the particle is given by:
$a = \frac{F(x)}{m} = 0$

Therefore, the acceleration of the particle for $|x| > X_0$ is zero.

Correct Option: D