A particle moves along the x-axis such that its velocity v varies with position x as v = alpha * sqrt(x), where alpha is a constant. If the particle starts from rest at x = 0 at t = 0, what is the acceleration of the particle at any time t?

A. ^22 — The acceleration is determined by differentiating the velocity function with respect to position and applying the chain rule.

JEE Main · PhysicsHard

$A particle moves along the x-axis such that its velocity v varies with position x as v = α x, where α is a constant. If the particle starts from rest at x = 0 at t = 0, what is the acceleration of the particle at any time t ?$

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Correct answer: A — $\frac{α ^{2}}{2}$

$Given the velocity function v = α x, we use the kinematic relation a = v \frac{d v}{d x} .$
$Differentiating v with respect to x, we get \frac{d v}{d x} = α \cdot \frac{1}{2 x} = \frac{α}{2 x} .$
$Substituting these into the acceleration formula, we find a = (α x) \cdot (\frac{α}{2 x}) = \frac{α ^{2}}{2} .$
$Hence the answer is (A).$