Re: Acceleraton and Deceleration Problem
- From: "Don1" <dcshead@xxxxxxxxxxx>
- Date: 24 Feb 2006 13:20:10 -0800
tadchem wrote:
Scenario:
You accelerate at 1 g (9.8 m/s/s) to 25 km/h (6.94 m/s), maintain that,
and then decelerate at 1 g to rest, stopping exactly where required.
You maintain maximum allowable speed most of the way.
You accelerate at 10 g (98 m/s/s) to 25 km/h (6.94 m/s), maintain that,
and then decelerate at 10 g to rest, stopping exactly where required.
You maintain maximum allowable speed even longer.
If you could accelerate *instantaneously* (infinite m/s/s) to 25 km/h,
maintain that and stop instantly when you got to the other end (ouch!)
you would maintain *maximum speed* over the distance for maximum time,
and thus achieve minimum transit time of 12 minutes (= 5 km / 25 km/h =
0.2 hour).
See where this is going?
The required acceleration for minimum time is infinite, and would be so
*whatever* the maximum speed is.
Tom Davidson
Richmond, VA
Let it be known from the outset here: There is NO instantaneous
(infinite) acceleration. All motion is a change in position DURING
TIME.
If at least a couple of youse guys had paid attention to your algebra
lessons, you could have learned the equation for _average_
acceleration: a=vi-vt/(2t); not a=vi-vt/t; which is called
"instantaneous acceleration"; of which it is not.
Then you could have taught each other how to solve these sorts of
things.
.
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