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(C) Given that the force $(F)$ is inversely proportional to the speed $(v)$,we have $F = \frac{k}{v}$,where $k$ is a constant.Using Newton's second la

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linearly related to time

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(C) Given that the force $(F)$ is inversely proportional to the speed $(v)$,we have $F = \frac{k}{v}$,where $k$ is a constant.Using Newton's second law,$F = ma = m \frac{dv}{dt}$.Substituting $F$,we get $m \frac{dv}{dt} = \frac{k}{v}$.Rearranging the terms,we get $mv \, dv = k \, dt$.Integrating both sides,we get $\int mv \, dv = \int k \, dt$,which gives $\frac{1}{2} mv^2 = kt + C$.Assuming the body starts from rest at $t = 0$,the constant $C = 0$.Thus,the kinetic energy $(KE = \frac{1}{2} mv^2)$ is equal to $kt$.Therefore,the kinetic energy is linearly related to time.

If the force acting on a body is inversely proportional to its speed,then its kinetic energy is

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