Distance moved as a function of time

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SUMMARY

The discussion focuses on the derivation of the distance moved by a particle under constant power, specifically showing that the distance is proportional to t^(3/2). The derivation begins with the relationship P=Fv, leading to the equation v^2=(2P/m)t. From this, the velocity function is established as dx/dt=(2P/m)^(1/2)t^(1/2), which integrates to yield the distance function x=(2/3)(2P/m)t^(3/2). This analysis assumes initial conditions of x=0 and v=0 at t=0.

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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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