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(D) Power $(P)$ is defined as the rate of doing work, $P = Fv$. Since $P$ is constant, $Fv = 	ext{constant}$.Using Newton's second law, $F = ma = m \

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$\frac{3}{2}$

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(D) Power $(P)$ is defined as the rate of doing work, $P = Fv$. Since $P$ is constant, $Fv = 	ext{constant}$.Using Newton's second law, $F = ma = m \frac{dv}{dt}$.Thus, $m \frac{dv}{dt} v = P$, which implies $v dv = \frac{P}{m} dt$.Integrating both sides, $\int v dv = \int \frac{P}{m} dt$, we get $\frac{1}{2} v^2 = \frac{P}{m} t + C$.Assuming the body starts from rest at $t=0$, $C=0$, so $v^2 = \frac{2P}{m} t$, or $v = \sqrt{\frac{2P}{m}} t^{1/2}$.Since $v = \frac{ds}{dt}$, we have $\frac{ds}{dt} = k t^{1/2}$ where $k = \sqrt{\frac{2P}{m}}$.Integrating with respect to time, $s = \int k t^{1/2} dt = k \frac{t^{3/2}}{3/2} = \frac{2k}{3} t^{3/2}$.Therefore, $s \propto t^{3/2}$, which means $x = \frac{3}{2}$.

$A$ body is moving along a straight line under the influence of a constant power source. If the relation between the displacement $(s)$ of the body and time $(t)$ is $s \propto t^x$, then $x=$

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