A body is moved from rest along a straight li | vedclass

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(A) Given that power $P$ is constant.From the work-energy theorem,the work done by the machine in time $t$ is $W = P t$.Since the body starts from res

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(A) Given that power $P$ is constant.From the work-energy theorem,the work done by the machine in time $t$ is $W = P t$.Since the body starts from rest,$W = \Delta K.E. = \frac{1}{2} m v^2$.Equating the two,$P t = \frac{1}{2} m v^2$,which gives $v = \sqrt{\frac{2 P}{m}} t^{1/2}$.We know that $v = \frac{d s}{d t}$,so $\frac{d s}{d t} = \sqrt{\frac{2 P}{m}} t^{1/2}$.Integrating with respect to $t$ with initial condition $s(0) = 0$:$s = \int_0^t \sqrt{\frac{2 P}{m}} t^{1/2} dt = \sqrt{\frac{2 P}{m}} \left[ \frac{t^{3/2}}{3/2} \right]_0^t = \sqrt{\frac{2 P}{m}} \cdot \frac{2}{3} t^{3/2}$.Now,calculate the ratio $s/v$:$\frac{s}{v} = \frac{\sqrt{\frac{2 P}{m}} \cdot \frac{2}{3} t^{3/2}}{\sqrt{\frac{2 P}{m}} t^{1/2}} = \frac{2}{3} t$.This shows that $t = \frac{3}{2} (s/v)$,which is a linear equation of the form $y = m x$ where $y = t$ and $x = s/v$.Therefore,the graph of $t$ versus $s/v$ is a straight line passing through the origin.

$A$ body is moved from rest along a straight line by a machine delivering constant power. The ratio of displacement and velocity $(s/v)$ varies with time $t$ as :

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