Time to Stop for Decelerating Particle – Rankers Physics

Calculus Based Questions: Practice Problem & Solution

A point moves linearly such that \(\frac{dv}{dt} = -\alpha v^{1/2}\), where \(\alpha\) is a positive constant. At start, \(v = v_0\). Find time taken by particle to stop?
\(\frac{\sqrt{v_0}}{\alpha}\)
\(\frac{v_0}{\alpha}\)
\(\frac{2\sqrt{v_0}}{\alpha}\)
\(\alpha\sqrt{v_0}\)

Solution Explained:

To solve this problem, we apply the core principles of Calculus Based Questions. Understanding the underlying formula is key to arriving at the correct answer below:

Integrating \(v^{-1/2} dv = -\alpha dt\) from \(v_0\) to \(0\) gives \(2\sqrt{v_0} = \alpha t\), so the time taken to stop is \(t = \frac{2\sqrt{v_0}}{\alpha}\).

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