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

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Kinematics

A particle initially at rest starts moving from reference point

x = 0

along x-axis, with velocity

v

that varies as

v = 4 x m / s

. The acceleration of the particle is ____

ms^{- 2}

Answer: 8

Solution

Using

a = v \frac{d v}{d x}

with

v = 4 x

\frac{d v}{d x} = \frac{2}{x}

, so

a = 4 x \times \frac{2}{x} = 8 m/s^{2}

, independent of x.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer

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