Question
A point mass m moves in a straight line under a retardation kv2 [where k is a positive constant and v is the instantaneous velocity]. The initial velocity of the point mass is u. The displacement of the point mass at time t is :
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Solution
[Key Concept: Under a retardation proportional to v2, 1/v grows linearly with time, so displacement grows logarithmically.]
$${a = - {kv}^{2} }\begin{matrix} & \frac{dv}{dt} = - {kv}^{2} \\ & \ \int_{u}^{v}\mspace{2mu}\mspace{2mu}\frac{dv}{v^{2}} = - k\int_{0}^{t}\mspace{2mu}\mspace{2mu} dt \\ & \ - \left. \ \frac{1}{v} \right|_{0}^{v} = - kt \\ & \frac{1}{v} - \frac{1}{u} = kt \\ & \frac{1}{v} = \frac{1}{u} + kt \\ & \ \int_{0}^{x}\mspace{2mu}\mspace{2mu} dx = \int_{0}^{t}\mspace{2mu}\mspace{2mu}\frac{u}{1 + ukt}dt \\ & x = u\frac{\left. \ \mathcal{l}n(1 + ukt) \right|_{0}^{t}}{uk} \\ & x = \frac{1}{k}ln(1 + ukt) \end{matrix}$$
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