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Question
hi.
x(t)=x(0)+v(0)t+ 1/2a t^2
does this come from the taylor formula
that f(x)= f(0) + f'(0) *x /1! + f''(0) *x^2 /(2!)?
just for information I guess. And if so then we don't need to make the long of proof of using a= deltav /t
and delta D= v(avg)*t to get this formula. second question is even if acceleration is not constant can we still use this formula by making it continue with f'''(x)*x^3 /3! until the n for which d^n x(t)/dt^n =C ?
Hi Hamad,
no, the kinetic formula does not come from the Taylor expansion. The kinetic formula is exact, while Taylor expansion limited to a finite number of terms is approximate. The kin. formula comes precisely from the assumption of a constant acceleration, i.e. d^2(x)/dt^2=a=const. From this we get the velocity formula v(t) = v0 + at first by one integration and the position formula x(t)=x0 + v0t + at^2/2 by another integration. This double integration is all the proof we need. The reason for it to look like the Taylor expansion is simply that Taylor expansion is polynomial and motion with a constant acceleration is quadratic, which is a polynomial.
For movement with non-constant acceleration of a known functional dependence on space and time d^2(x)/dt^2 = a = a(x,y,z,t) we have to solve this second order ordinary diff. equation, Taylor expansion is of no use.
Cheers,
Daniel
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Daniel Mazur
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