ExamBro
ExamBro
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 _______ \( \mathrm{ms}^{-2}\).

  1. A \(7\)
  2. B \(8\)
  3. C \(9\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(8\)

Step-by-step Solution

Detailed explanation

\(\mathrm{V}=4 \sqrt{\mathrm{x}}\) \(\mathrm{a}=\mathrm{V} \frac{\mathrm{dv}}{\mathrm{dx}}\) \(=4 \sqrt{\mathrm{x}} \times 4 \times \frac{1}{2} \mathrm{x}^{-1 / 2}=8 \mathrm{~m} / \mathrm{s}^2\)
Same subject
Explore more questions on app