Acceleration Proportionality – Rankers Physics
Topic: Kinematics
Subtopic: Calculus Based Questions

Acceleration Proportionality

A particle moves a distance (x) in time (t) according to equation \(x = (t + 5)^{-1}\). The acceleration of particle is proportional to:

(2010 Pre)

\(\text{Velocity}^{2/3}\)
\(\text{Velocity}^{3/2}\)
\((\text{Distance})^2\)
\((\text{Distance})^{-2}\)

Solution:

Given \(x = (t + 5)^{-1}\). Velocity \(v = \frac{dx}{dt} = -(t + 5)^{-2}\). Acceleration \(a = \frac{dv}{dt} = 2(t + 5)^{-3}\). From \(v = -(t + 5)^{-2}\), we have \((t+5)^{-1} = ((-v)^{-1/2})\). So \(a = 2((t+5)^{-1})^3 = 2((-v)^{-1/2})^3 = 2(-v)^{3/2}\). Thus, \(a \propto (\text{Velocity})^{3/2}\).

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