Question:
A particle initially at rest starts moving from origin along X - axis with velocity 'V' that varies as $V = 2\sqrt{x}~ms^{-1}$. The acceleration of the particle in $ms^{-2}$ is:
A particle initially at rest starts moving from origin along X - axis with velocity 'V' that varies as $V = 2\sqrt{x}~ms^{-1}$. The acceleration of the particle in $ms^{-2}$ is:
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If $v \propto \sqrt{x}$, acceleration is constant.
Updated On: Jun 6, 2026
- 2
- $2\sqrt{2}$
- $2/\sqrt{2}$
- Zero
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Kinematics: Acceleration formula $a = v \frac{dv}{dx}$.
Step 2: Meaning
Differentiating velocity with respect to position to find acceleration.
Step 3: Analysis
Given $V = 2x^{1/2}$. $\frac{dV}{dx} = 2 \cdot \frac{1}{2} x^{-1/2} = \frac{1}{\sqrt{x}}$. $a = V \frac{dV}{dx} = (2\sqrt{x}) \cdot (\frac{1}{\sqrt{x}}) = 2~ms^{-2}$.
Step 4: Conclusion
The acceleration is a constant 2.
Final Answer: (A)
Kinematics: Acceleration formula $a = v \frac{dv}{dx}$.
Step 2: Meaning
Differentiating velocity with respect to position to find acceleration.
Step 3: Analysis
Given $V = 2x^{1/2}$. $\frac{dV}{dx} = 2 \cdot \frac{1}{2} x^{-1/2} = \frac{1}{\sqrt{x}}$. $a = V \frac{dV}{dx} = (2\sqrt{x}) \cdot (\frac{1}{\sqrt{x}}) = 2~ms^{-2}$.
Step 4: Conclusion
The acceleration is a constant 2.
Final Answer: (A)
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