in x = x0 + v t you must have assumed v0 t0 = 0

Subject:	in x = x0 + v t you must have assumed v0 t0 = 0
Date:	Thu, 3 Dec 2009 08:02:29 -0800 (PST)

and if so x = x0 + v t + v0 t0 breaks down to [x = v t]_{by any
index}. you were saying Galileo's and Einstein's relativity? well come
again please but skip Newton's ill differential mechanics! after all
in the beginning x cannot be function that depends only on time (say x
= 1 + 2 t + 3 t t; the adders left are not even addable among them
self) so in the end you can claim x = x0 + v0 t + a t t / 2 = function
(x0, v0, a, t). if some magnitude becomes constant it doesn't prevent
the function to continue depending on its unique units. try plotting x
= v t in wolfram's matematica:

http://www.wolframalpha.com/input/?i=Plot[v+t%2C+{v%2C+-4%2C+%2B4}%2C+{t%2C+-4%2C+%2B4}]

lesson: no more shitty kinetics right? find your self another way of
doing physics...

generated at 02:27:57