A particle initially at rest starts moving from reference point. along x-axis, with velocity that varies as m/s. The acceleration of the particle is

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Solution

Q.
A particle initially at rest starts moving from reference point. $x = 0$ along x-axis, with velocity $v$ that varies as $v = 4 \sqrt{x}$ m/s. The acceleration of the particle is _____ $m s^{- 2}$ . [2024]

Ans.
(8) Given, $v = 4 \sqrt{x}$

   $‐ ‐ ‐ □ ‐ ‐ ‐ \to ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ x - a x i s$

   $x = 0, t = 0$

   $a = \frac{v d v}{d x} = (4 \sqrt{x}) \frac{d (4 \sqrt{x})}{d x} = 16 \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}$

   $16 \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} = 8 {m / s}^{2}$

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