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

<p>To find the acceleration of the particle, we first need to differentiate the velocity function with respect to time. The velocity function given is</p> <p>$v = 4\sqrt{x}$</p> <p>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&#39;ll need to use the chain rule to differentiate $v$ with respect to $t$.</p> <p>The chain rule in this context can be stated as follows: </p> <p>$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$</p> <p>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 $\frac{dv}{dx}$:</p> <p>$v = 4\sqrt{x} = 4x^{\frac{1}{2}}$</p> <p>Differentiating with respect to $x$, we get:</p> <p>$$ \frac{dv}{dx} = 4 \cdot \frac{1}{2} x^{-\frac{1}{2}} = 2x^{-\frac{1}{2}} = \frac{2}{\sqrt{x}} $$</p> <p>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:</p> <p>$\frac{dv}{dx} = \frac{2}{\sqrt{x}} = \frac{2}{\frac{v}{4}} = \frac{8}{v}$</p> <p>Now, using the chain rule:</p> <p>$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{8}{v} \cdot v$</p> <p>Simplifying this, the velocity terms cancel out, leaving us with:</p> <p>$a = 8 \text{ ms}^{-2}$</p> <p>Thus, the acceleration of the particle is $8 \text{ ms}^{-2}$.</p>