QJEEM24Phcb6d: JEE Main 2024 Physics - Class 11 JEEM24Phcb6d Solution: A particle initially at rest starts moving from reference point …

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:

$a = 8 \text{ ms}^{-2}$

Thus, the acceleration of the particle is $8 \text{ ms}^{-2}$ .