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

Correct Answer: 8

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 \( 8 \, \mathrm{ms^{-2}} \).