To find the acceleration of the particle, we first need to differentiate the velocity function with respect to time. The velocity function given is

$$ v = 4\sqrt{x} $$

However, this function gives the velocity as a function of position $x$, not as a function of time $t$. Since acceleration is the rate of change of velocity with respect to time, we'll need to use the chain rule to differentiate $v$ with respect to $t$.

The chain rule in this context can be stated as follows:

$$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} $$

Now, because $\frac{dx}{dt}$ is the velocity $v$ itself and $\frac{dv}{dx}$ is the derivative of the velocity with respect to $x$, we first find $\frac{dv}{dx}$:

$$ v = 4\sqrt{x} = 4x^{\frac{1}{2}} $$

Differentiating with respect to $x$, we get:

$$ \frac{dv}{dx} = 4 \cdot \frac{1}{2} x^{-\frac{1}{2}} = 2x^{-\frac{1}{2}} = \frac{2}{\sqrt{x}} $$

Now, because $v = 4\sqrt{x}$, we can rewrite $\sqrt{x}$ as $\frac{v}{4}$. Using this to replace $\sqrt{x}$ in our expression for $\frac{dv}{dx}$, we get:

$$ \frac{dv}{dx} = \frac{2}{\sqrt{x}} = \frac{2}{\frac{v}{4}} = \frac{8}{v} $$

Now, using the chain rule:

$$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{8}{v} \cdot v $$

Simplifying this, the velocity terms cancel out, leaving us with:

$$ a = 8 \text{ ms}^{-2} $$

Thus, the acceleration of the particle is $8 \text{ ms}^{-2}$.