Question 104: Motion in 1D

Question

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 __________ $\mathrm{ms}^{-2}$.

Accepted Answer

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:
$$ a = 8 	ext{ ms}^{-2} $$
Thus, the acceleration of the particle is $8 	ext{ ms}^{-2}$.

Motion in 1D

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}.

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