A particle initially at rest starts moving from reference point. x =0 along x -axis, with velocity v that variesas v=4√(x) m / s . The acceleration of the particle is ms -2

Physics · Motion in one Dimension

JEE Main 2024 — 1 February, Shift 2 — Question 52

A particle initially at rest starts moving from reference point. $\mathrm{x}=0$ along x -axis, with velocity $v$ that variesas $v=4\sqrt{x}\mathrm{~m}/\mathrm{s}$ . The acceleration of the particle is $\qquad$ $\mathrm{ms}{-2}$

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

$V=4 \sqrt{x}$ $a=V \frac{d v}{d x}$ $=4 \sqrt{\mathrm{x}} \times 4 \times \frac{1}{2} \mathrm{x}^{-1 / 2}=8 \mathrm{~m} / \mathrm{s}^{2}$

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Physics
Chapter: Motion in one Dimension
Topic: Non-Uniformly Accelerated Motion