A particle initially at rest starts moving from reference point $$x = 0$$ along x-axis, with velocity $$v$$ that varies as $$v = 4\sqrt{x}$$ m s$$^{-1}$$. The acceleration of the particle is ______ m s$$^{-2}$$.

Find the acceleration of a particle with velocity $$v = 4\sqrt{x}$$ m/s.

Acceleration can be expressed as $$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v $$ because by the chain rule, $$\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$$ and $$\frac{dx}{dt} = v$$.

Since $$v = 4\sqrt{x} = 4x^{1/2}$$, it follows that $$ \frac{dv}{dx} = 4 \times \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x}} $$.

Substituting into the expression for acceleration gives $$ a = \frac{dv}{dx} \cdot v = \frac{2}{\sqrt{x}} \times 4\sqrt{x} = \frac{2 \times 4\sqrt{x}}{\sqrt{x}} = 8 \text{ m/s}^2 $$. Note that $$\sqrt{x}$$ cancels, making the acceleration constant and independent of position.

The answer is 8 m/s$$^2$$.