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A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time 't' is proportional to

\[(a)\;t^{\large\frac{1}{2}} \quad (b)\;t^{\large\frac{3}{2}} \quad (c)\;t^{\large\frac{3}{4}} \quad (d)\;t^2\]

1 Answer

Answer: $t^{\large\frac{3}{2}}$
Power $ P = F\;v = ma\; v$
Acceleration $a = \large\frac{S}{t^2}$ and Velocity $v = \large\frac{S}{t}$, where $S$ is the distance moved.
Since the Power $P$ and mass $m$ are constant, we get
$P = \large\frac{mS}{t^2}$$\times \large\frac{S}{t}$$\rightarrow S^2 = t^3 \rightarrow S \propto t^{\large\frac{3}{2}}$
answered Jul 17, 2013 by meena.p
edited Aug 25, 2014 by balaji.thirumalai

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