In article <xSl19.3937$Py1.69178 at news2.ulv.nextra.no> "Thalamus" <zhil at online.no> writes:
<"Cary Kittrell" <cary at afone.as.arizona.edu> skrev i melding
<news:ai4m4i$q2n$1 at oasis.ccit.arizona.edu...
<> In article <8ko09.28058$Fq6.2865116 at news2.west.cox.net> "John Knight"
<<johnknight at usa.com> writes:
<> Only if you believe that:
<>
<> S = at^2
<>
<> (check any physics book; he was wrong)
<
<It was like this, retard:
<
<F=m(v/t) - a=(v-v0)/t - a is acceleration,v is velocity, t is time, F is
<force
<
<t=mv/F - exchanging t with F, see the likeness of the equations ??
<
<t=mv/ma - insert ma instead of F (F=ma)
<t=v/a - shorten the thing, by dispatching off with m (mass).
<t=(S/t)/a - here's the tricky part, insert S/t instead of v (S=vt or in my
<opinion v=S/t).
<t=(S/ta) - shorten the whole thing, so it is elegant.
<t²=(S/a) - transfer t to one side of the equation, and voila !!
<t=sqr(S/a) - you have Brian's equation of time, height and acceleration.
Yep, that's what you get, all right: Brian's equation. Unfortunately,
Mr. Newton's equation differs from yours by a factor of two, as I
originally pointed out.
<
<You loose, I win - as I am a Superior White God, and you're just a silly
<feminine creature :-)
<
Sorry, SWG, but this silly feminine creature realizes that
there's an implicit assumption of linearity in your step 5, where
you substitute S/t for v. That's true only for uniform velocity;
it's not true under acceleration, where velocity is constantly
increasing. In that case you can't do it (in a straightforward
manner) with algebra, you have to use calculus. In particular,
you have to integrate:
dS/dt = a*t, or
dS = integral (a*t*dt)
the solution to which is, of course, 1/2 at^2, not at^2. Which
is what I said originally. You fall a mile in 18 seconds, not
two miles.
As I said to John, check any physics book. Or if you're just
too lazy, here's the first of a roughly a zillion hits on the net:
http://c3po.lpl.arizona.edu/~jbarnes/nats102/HW2/
-- cary
