Distance moved as a function of time

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Amith2006
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Sir,
A body is moved along a straight line by a machine delivering constant power. It is said that the distance moved by the particle in time t is proportional to t^(3/2). Can you please explain how this is derived?
Here the symbol ^ represents power.
 
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P=Fv=v m dv/dt=(m/2)d/dt(v^2), so
v^2=(2P/m)t.
Then, dx/dt=(2P/m)^1/2 t^1/2,
and x=(2/3)(2P/m)t^3/2.
This assumes x=-0 and v=0 agt t=0.
 
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The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
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