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. Numerical answer type. Work it out, then open the answer.

Answer: 8. v² = 16x a = vdv/dx = 1/2(d(v²))/(dx) = 16/2 So a = 8 m/s².

A particle initially at rest starts moving from reference point x = 0… | JEE Main 2024 (1 Feb, Shift 2)

Q13JEE Main 2024 · 1 FEB · SHIFT 2Medium

Velocity, Acceleration & Variable Motion

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

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 8

$v^2 = 16x$
$a = v\dfrac{dv}{dx} = \dfrac{1}{2}\dfrac{d(v^2)}{dx} = \dfrac{16}{2}$
So $a = 8\text{ m/s}^2$ .

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