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Re: acceleration

It's worse than ambiguous - it's discontinuous at the peak of a vertical
trajectory. It goes instantaneously from a constant negative value to a
constant positive value near the peak. I'm not sure what the point is
here.

Bob at PC

"John S. Denker" wrote:

I wrote:
>>
>> Asking what is "the" acceleration at the peak of a
>> parabolic arc is at best ambiguous.
>> -- The scalar acceleration is zero. The speed
>> is locally constant at this point.
>> -- The vector acceleration is of course just g.

On 11/19/2003 06:47 PM, cliff parker wrote:
>
> At what point isn't the speed locally constant?

At all other points (other than the peak of the trajectory)
the speed is not locally constant. "Locally" is shorthand
for "to first order".

To see this, calculate (d/dt)|v|. If you're feeling
lazy, look it up e.g.
http://mathworld.wolfram.com/Acceleration.html
(where beware s denotes arc_length, not speed).
Equation 6 is particularly elegant and easy to interpret:
(scalar acceleration) = (unit tangent) dot (vector acceleration)