How Do I Solve Part B of the Lennard-Jones Potential Problem?

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The discussion centers on solving part B of a problem related to the Lennard-Jones potential. The user contemplates using the relationship T = E - U to find velocity (v) and integrate for position (x), questioning whether the constant energy (E) can be ignored. They derive an expression for x but arrive at an unexpected result for the minimum point, r, by setting the derivative of the potential (dV/dr) to zero. The approach involves simplifying the problem by canceling constants in the derivation process. The user seeks clarification on the treatment of energy in their calculations.
Oblio
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I'm really stuck on b.) of this question.

http://www.physics.brocku.ca/Courses/2P20/problems/prob_lenardjones.pdf

I think I can use T = E - U, and solve for v from within T and integrate for x...

but what do I do with E?
 
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can i ignore E since its a constant? I got something like,
x = /frac{-6r^6A + 12B}{2r^13} t^2 [\tex]
 
The minimum is the point at which dV/dr=0
 
I'm getting the ridiculous value of :

r = \sqrt[6]{\frac{-12B}{6A}}

by deriving the two quotients, allowing A' and B' to cancel in the process (constants) cancelling out the high denominators as much as possible... etc..
 

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