Velocity-Displacement Relation – Rankers Physics
Topic: Kinematics
Subtopic: Calculus Based Questions

Velocity-Displacement Relation

A particle begins to move along straight line where the acceleration \( (a) \) of the particle varies with displacement \( (x) \) according to relation, \( a = 5x \), then velocity of the particle varies with displacement as
\( x^{1/2} \)
\( x^1 \)
\( x^{1/3} \)
\( x^{3/4} \)

Solution:

Using \( a = v \frac{dv}{dx} \), we write \( v \frac{dv}{dx} = 5x \). Integrating both sides, \( \int v \, dv = \int 5x \, dx ⇒ \frac{v^2}{2} = \frac{5x^2}{2} + C \). Assuming the particle starts from rest, \( v^2 \propto x^2 ⇒ v \propto x^1 \).

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