How to derive x(t) equation from energy

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The discussion focuses on deriving the equation x(t) from energy principles in classical mechanics. The integral equation presented, \int_{x_0}^x \frac{dx}{\pm \sqrt{{\frac{2}{m}\{E-V\left( x\right)\}}}} = t - t0, is analyzed with respect to potential energy V(x) and total energy E. The key conclusion is that using initial conditions v0 and x0 instead of final values leads to the correct formulation of the equation x = x0 + v0t + at²/2. The integration process requires careful handling of constants and initial conditions to achieve the desired result.

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1. Using V(x)= -max, in the following equation:
[tex]\int_{x_0}^x \frac{dx}{\pm \sqrt{{\frac{2}{m}\{E-V\left( x\right)\}}}}<br /> \[/tex] = t - t0

to get:
x = x0 + v0 + at2/2

E is total energy and V(x) is potential energy. I have tried hard integrating it in various ways but do not seem to get the required result.
I would really appreciate in help or tips in this regard.When I use E - 0.5mv^2= V(x), the denominator becomes v and really does not help at all. If I do not do that, and use V(x) = -max that does not help either. I do not seem to be reaching the required equation in any way.
 
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hi cream3.14159! welcome to pf! :smile:

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cream3.14159 said:
… to get:
x = x0 + v0 + at2/2

When I use E - 0.5mv^2= V(x)

(it should of course be x = x0 + v0t + at2/2)

why are you using E - 0.5mv2 ? :confused:

this is a perfectly ordinary integral of (constant - 2ax)-1/2

show us what you get :smile:
 
Hi!

Thank you for the response. I solved it with someone's help. The mistake I was doing was to use 0.5mv2-m*a*x to replace E. However, using v0 and x0 instead of v and x in this expression works to give the desired result and also, one has to put t0 = 0.
 

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