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JEE Main 20241 FebruaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 1 February
Shift: Evening Shift
Subject: Physics
Chapter: Motion in One Dimension

JEE Main 2024 · 1 February · Evening Shift · Physics · Motion in One Dimension

A particle initially at rest starts moving from reference point x =...

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}^{- 1}

. The acceleration of the particle is _____

m{s}^{- 2}

.

Correct answer

0

Step-by-step solution

Given: $v = 4\sqrt{x}$

Instantaneous acceleration of the particle can be written as, $a = v\dfrac{dv}{dx}$

$= \left(4\sqrt{x}\right) \times \left(4 \times \dfrac{1}{2}{x}^{- 1 / 2}\right) = 8m{s}^{- 2}$

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