Posted by Abi on Friday, February 11, 2011 at 3:38pm.
which has greater acceleration at t near 0?
Alpha a = alpha/sqrt t
which is so big near 0 that it is undefined
whereas
beta is just an ordinaty old constant acceleration.
Try some of the Physics Tutorials:
http://search.yahoo.com/search?fr=mcafee&p=Physics+tutorial
Sra
SraJMcGin,
Do you see something incorrect about my reply?
Damon
v = integral 0 to t of a dt
va = alpha 2 t^(1/2)
vb =beta t
xa = alpha (2/3) t^(3/2)
xb = beta (1/2) t^2
when xa = xb
alpha (2/3) t^(3/2) = beta (1/2) t^2
t^(1/2) = [ (4/3)(alpha/beta)]
so
t^2 = [ (4/3)(alpha/beta)]^4
and x = (1/2) beta t^2 = (1/2)beta [ (4/3)(alpha/beta)]^4
Hi Damon,
Thanks for your help. Can you explain me how you got alpha (2/3) t^(3/2) ? Because I got
alpha (4/3) t^(3/2) = beta (1/2) t^2 and then t= [(8/3)(alpha/beta)]^2 I then used my equation of beta and substituted t with my answer and I got x=(4096aplpha^4)/(162beta^3)
if
va = alpha 2 t^(1/2)
then the integral of alpha 2 t^(1/2) dt
is
2 alpha * (2/3) t^(3/2)
= (4/3) alpha t^(3/2) You are right.
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