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 determine the particle's acceleration from the velocity function \( v = 4\sqrt{x} \, \text{m/s} \), we must compute the derivative of velocity with respect to time \( t \), yielding acceleration \( a \).

Step 1: 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: 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: Substitute this result into the acceleration equation:

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

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

Verification: This value is confirmed to be within the specified range \( 8, 8 \).