Is My Physics Homework Calculation Correct?

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SUMMARY

The discussion centers on the calculation of kinetic energy and its relationship with power in a physics homework problem. The user correctly derives the velocity equation as v = √(2Pt/m) for part a, establishing the connection between power, time, and mass. For part b, the user differentiates the velocity equation to find acceleration and integrates to determine distance, yielding s = (2/3)(√(2p/m))t^(3/2). The calculations are confirmed as accurate, provided the initial assumption regarding power and kinetic energy is correct.

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Homework Statement


http://img120.imageshack.us/img120/3743/physicsquestionnb9.jpg
http://g.imageshack.us/img120/physicsquestionnb9.jpg/1/


The Attempt at a Solution



I'm trying this problem and I'm really not sure if my answers are correct at all.

For part a) I get the kinetic energy = Pt (since it is the electrical energy)
I make 0.5m.v^2=Pt and make v the subject, leaving [itex]v=\sqrt \frac{2pt}{m}[/itex]

Now I hope what I've done so far is okay. For part b to find the acceleration I differentiated the equation I derived above. To do this I wrote the equation as
[itex]v=(\sqrt \frac{2p}{m})t^\frac{1}{2}[/itex]

and so

[itex]dv/dt = \frac{1}{2}(\sqrt \frac{2p}{m})t^\frac{-1}{2}[/itex]

While for the distance, by integration, I got

[itex]s= \frac{2}{3}(\sqrt \frac{2p}{m})t^\frac{3}{2}[/itex]


Im not very confident in my maths so I'd appreciated it if somebody could check whether I got the right results.
 
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As long as your first assumption is correct, that power times t is equal to the kinetic energy, you did the integration and the derivations correctly.
 

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