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Re: Problem



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.
( This case was excluded from the original question, which contemplated an
acceleration perpendicular to the velocity.)

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "Bob Sciamanda" <trebor@VELOCITY.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, September 18, 2001 4:41 PM
Subject: Re: Problem


----- Original Message -----
From: "Hugh Haskell" <hhaskell@MINDSPRING.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, September 18, 2001 4:21 PM
Subject: Re: Problem


At 15:12 -0400 9/18/01, Bob Sciamanda wrote:

Hugh, the square of the speed is given by

V(t)^2 = (Vo*CosTH)^2 +(Vo*SinTH-gt)^2

The time derivative of this goes through zero at the top of the
trajectory,
ie: at t = (Vo*SinTH)/g

I think not, Bob. If you take the derivative of V(t) (the square root
of your espression above), I think that you will find that it equals
-g, which it should, since this is in a uniform grav. field, by the
statement of the problem.

Hugh:
I get dV/dt =( -g/V )*(Vo*Sin TH -gt); where V is the speed.
This goes through zero at t=(Vo*SinTH)/g.

The derivative of the speed is NOT the same as the magnitude of the
acceleration except, at times, by coincidence!

Furthermore, take the special case when TH=90 degrees, i.e., a
vertical trajectory. Are you telling me that when the object gets to
the very top it's acceleration goes to zero? IF so, how does it get
back down?

Hugh,
The derivative of the speed is NOT the same as the magnitude of the
acceleration except, at times, by coincidence!

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor