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 ___ \(\text{ms}^{-2}\).
The acceleration of the particle is ___ \(\text{ms}^{-2}\).
Correct Answer: 8
Approach Solution - 1
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 \( 8, 8 \)
Approach Solution -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 \( 8 \, \mathrm{ms^{-2}} \).
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