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(C) The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$. Given that kinetic energy increases uniformly with time,we have $K = kt$,where $k$ is a

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$\frac{1}{\sqrt{t}}$

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(C) The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$. Given that kinetic energy increases uniformly with time,we have $K = kt$,where $k$ is a constant. Thus,$\frac{1}{2}mv^2 = kt$,which implies $v^2 \propto t$,or $v \propto t^{1/2}$. The acceleration $a$ is the rate of change of velocity: $a = \frac{dv}{dt} = \frac{d}{dt}(ct^{1/2}) = \frac{1}{2}ct^{-1/2}$. Therefore,$a \propto \frac{1}{\sqrt{t}}$. According to Newton's second law,the net force $F = ma$. Since $m$ is constant,$F \propto a$,which means $F \propto \frac{1}{\sqrt{t}}$.

At time $t=0$,a particle starts moving along the $x$-axis. If its kinetic energy increases uniformly with time $t$,the net force acting on it must be proportional to

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