Step 1: What is given.
A particle starts from the origin and its speed depends on position $x$ as $V = 2\sqrt{x}\ \text{m s}^{-1}$. We need its acceleration.
Step 2: The right formula.
When velocity is written as a function of position (not time), the neat way to get acceleration is to use the chain rule form. \[ a = V\,\frac{dV}{dx} \] This works because $a = \dfrac{dV}{dt} = \dfrac{dV}{dx}\cdot\dfrac{dx}{dt} = V\dfrac{dV}{dx}$.
Step 3: Differentiate the velocity.
Write $V = 2x^{1/2}$. Then \[ \frac{dV}{dx} = 2\cdot\tfrac{1}{2}x^{-1/2} = \frac{1}{\sqrt{x}} \]
Step 4: Put it together.
\[ a = V\,\frac{dV}{dx} = \left(2\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\right) = 2\ \text{m s}^{-2} \] The $\sqrt{x}$ terms cancel, so the acceleration does not depend on position at all.
Step 5: Conclusion.
The acceleration is a constant value of $2\ \text{m s}^{-2}$. \[ \boxed{2\ \text{m s}^{-2}} \]