JEE Mains · Physics · STD 11 - 2. motion in straight line
A particle initially at rest starts moving from reference point. \(\mathrm{x}=0\) along \(\mathrm{x}\)-axis, with velocity \(v\) that varies as \(v=4 \sqrt{\mathrm{x} m} / \mathrm{s}\). The acceleration of the particle is _______ \( \mathrm{ms}^{-2}\).
- A \(7\)
- B \(8\)
- C \(9\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(8\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}=4 \sqrt{\mathrm{x}}\) \(\mathrm{a}=\mathrm{V} \frac{\mathrm{dv}}{\mathrm{dx}}\) \(=4 \sqrt{\mathrm{x}} \times 4 \times \frac{1}{2} \mathrm{x}^{-1 / 2}=8 \mathrm{~m} / \mathrm{s}^2\)
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