Question

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

Answer 1

Given: - Velocity of the particle: $v = 4\sqrt{x} \, \mathrm{m/s}$ - The particle starts from rest at $x = 0$.

Step 1: Expressing Velocity as a Function of Position

The velocity $v$ is given by: $v = 4\sqrt{x}$ Squaring both sides: $v^2 = (4\sqrt{x})^2 = 16x$

Step 2: Using the Relation Between Acceleration, Velocity, and Position

The acceleration $a$ is given by: $a = v \frac{dv}{dx}$ Differentiating $v^2 = 16x$ with respect to $x$: $\frac{d(v^2)}{dx} = 16$ Using the chain rule: $2v \frac{dv}{dx} = 16$ Rearranging: $v \frac{dv}{dx} = 8$ Thus, the acceleration is: $a = 8 \, \mathrm{ms^{-2}}$

Conclusion: The acceleration of the particle is $8 \, \mathrm{ms^{-2}}$.