JEE MainPhysicsToughMotion in 1D

A point moves in a straight line under the retardation av^2,
where is a positive constant and v is speed. If the initial speed
is u, the distance covered in seconds is

Show answer and solution

The retardation is given by

\frac{dv}{dt} = - av^{2}

integrating between proper limits

\Rightarrow - \int_{u}^{v}\mspace{2mu}\frac{dv}{v^{2}} = \int_{0}^{t}\mspace{2mu} adt

\frac{1}{v} = at + \frac{1}{u}

\Rightarrow \frac{dt}{dx} = at + \frac{1}{u} \Rightarrow dx = \frac{udt}{1 + aut}

integrating between proper
limits

\Rightarrow \int_{0}^{s}\mspace{2mu} dx = \int_{0}^{t}\mspace{2mu}\frac{udt}{1 + aut} \Rightarrow S = \frac{1}{a}\ ln\ (1 + \ aut\ )

Source: Grass-G / P-05 Physics paper 10 October.docx

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