Solution:
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 \(t = \frac{2\sqrt{v_0}}{\alpha}\).
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?
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