1. We are given the velocity as a function of position:

$v = 4\sqrt{x}$

We need to find the acceleration $a$.

2. Use the relation between acceleration, velocity, and position:

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

3. Differentiate $v$ with respect to $x$:

$v = 4x^{1/2}$

So,

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

4. Now substitute into

$a = v\frac{dv}{dx}$

$a = \left(4\sqrt{x}\right)\left(\frac{2}{\sqrt{x}}\right)$

$a = 8$

5. Therefore, the acceleration is constant:

$\boxed{8\ \text{m s}^{-2}}$