Re: Problem

[Bob Sciamanda <trebor@VELOCITY.NET>]

You're absolutely right, John.  I was unconsciously thinking only of the
"absolute value" of the derivative of the speed.  Although the speed is
inherently positive, its time derivative IS SIGNED, as I neglected to
acknowledge.  Thanks (again)!

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "John Mallinckrodt" <ajmallinckro@CSUPOMONA.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, September 18, 2001 5:11 PM
Subject: Re: Problem


> On Tue, 18 Sep 2001, Bob Sciamanda wrote:
>
> > Hugh,
> > I have to take back my answer to your second question, about the TH=90
deg
> > case.  One dimensional motion is a special case where the derivative
of
> > the speed IS numerically equal to the magnitude of the acceleration,
and
> > in vertical free fall this is a constant (g), as you say.
>
> Hugh and Bob,
>
> Stop; you're both wrong.  Certs is *not* a ...   Er, that is, I
> mean ...
>
> As my recent posts pointed out, in the case of a vertically
> launched projectile, d|v|/dt = -g on the way up and d|v|/dt = +g
> on the way down.  At the top, d|v|/dt is undefined.
>
> John Mallinckrodt      mailto:ajm@csupomona.edu
> Cal Poly Pomona          http://www.csupomona.edu/~ajm