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(C) The power $P$ supplied to the particle is constant. Since power is the rate of change of kinetic energy,we have $\frac{dK}{dt} = P$. Integrating t

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(C) The power $P$ supplied to the particle is constant. Since power is the rate of change of kinetic energy,we have $\frac{dK}{dt} = P$. Integrating this with respect to time,we get $K = Pt$ (assuming initial kinetic energy is zero).Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = Pt$,which implies $v = \sqrt{\frac{2P}{m}} t^{1/2}$.Since velocity $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = \sqrt{\frac{2P}{m}} t^{1/2}$.Integrating with respect to time,$s = \int \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2}$.Thus,$s \propto t^{3/2}$.The graph of $s$ versus $t$ for $s \propto t^{3/2}$ is a curve that is concave upwards (increasing slope),which corresponds to Graph $C$.

$A$ particle is moving unidirectionally on a horizontal plane under the action of a constant power supplying energy source. The displacement $(s)$ - time $(t)$ graph that describes the motion of the particle is (graphs are drawn schematically and are not to scale)

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