Question:

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

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Constant power means \(Pt = \frac12 m v^2\), so \(v \propto t^{1/2}\); integrate once more to get the distance.
Updated On: Nov 4, 2026
  • \(t^{\frac{3}{2}}\)
  • \(t^{\frac{1}{2}}\)
  • \(t^{\frac{1}{4}}\)
  • \(t^{\frac{3}{4}}\)
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The Correct Option is A

Solution and Explanation

Work done by the machine in time \(t\): \(W = Pt\). Starting from rest, this equals the kinetic energy: \(Pt = \dfrac12 m v^2\).
\(v = \sqrt{\dfrac{2P}{m}}\;t^{1/2}\).
Distance: \(s = \displaystyle\int_0^t v\,dt = \sqrt{\dfrac{2P}{m}}\cdot\dfrac{2}{3}\,t^{3/2}\).
Hence \(s \propto t^{3/2}\).
Answer: (1) \(t^{3/2}\)
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