Since velocity is a time rate of change in position [s/t], and applies at a
particular instant _point in time_ ; then a time rate of change in velocity
at a particular point in time, must be a time rate of change; in a time rate
of change in position [a = (vt-vi)/t = 2s/t²]; so that: [a/2 = (vt-vi)/(2t)
= s/t²]:

Galileo found that all objects; bodies, and masses thereof, fell with an
ever increasing impetus, and/or momentum of about s/t² = 16'/sec²;
regardless of their weight: So that the distance [s] that they fell after a
period of [t] seconds, could be calculated as: [s = (16'/sec²)t²].

Using today's precision instruments, we can more accurately calculate this;
to the nth decimal place.

Subject: Time rate of change in velocity
From: "Donald G. Shead"
Date: 01/09/03 19:27 GMT Daylight Time
Message-id:

Since velocity is a time rate of change in position [s/t], and applies at a
particular instant _point in time_ ; then a time rate of change in velocity
at a particular point in time, must be a time rate of change; in a time rate
of change in position [a = (vt-vi)/t = 2s/t²]; so that: [a/2 = (vt-vi)/(2t)
= s/t²]:

Galileo found that all objects; bodies, and masses thereof, fell with an
ever increasing impetus, and/or momentum of about s/t² = 16'/sec²;
regardless of their weight: So that the distance [s] that they fell after a
period of [t] seconds, could be calculated as: [s = (16'/sec²)t²].

Using today's precision instruments, we can more accurately calculate this;
to the nth decimal place.


and your question is ?

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"Donald G. Shead" wrote:

Since velocity is a time rate of change in position....

Acceleration, a = dv/dt
http://scienceworld.wolfram.com/phys...eleration.html

"Donald G. Shead" wrote:

Since velocity is a time rate of change in position [s/t], and applies at a
particular instant _point in time_ ;
[snip]

Hey ****Head - calculus.

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