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

To find the acceleration of a particle from the given velocity function \( v = 4\sqrt{x} \, \text{m/s} \), we need to calculate the derivative of velocity with respect to time \( t \), which will give us the acceleration \( a \). 

Step 1: First, express velocity in terms of position using the chain rule:

\( v = \frac{dx}{dt} \) and \( a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \)

Step 2: Substitute \( v \) for \( \frac{dx}{dt} \):

\( a = v \cdot \frac{dv}{dx} \)

Step 3: Next, differentiate \( v = 4\sqrt{x} \) with respect to \( x \):

\[ \frac{dv}{dx} = \frac{d}{dx} \left( 4\sqrt{x} \right) = 4 \cdot \frac{1}{2\sqrt{x}} = \frac{2}{\sqrt{x}} \]

Step 4: Substituting this back into the expression for acceleration:

\( a = (4\sqrt{x}) \cdot \frac{2}{\sqrt{x}} = 8 \)

Final Answer: Thus, the acceleration of the particle is \( 8 \, \text{m/s}^2 \).

Verification: Verify that this value falls within the given range \( 8, 8 \)

Answer 2

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