Free fall
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From: odin on 16 Sep 2005 01:53 > Look in your basic physics text: Although they probably use s instead > of d; the equation is given as s=(vi)t + at^2; No wonder it only works > when vi=0'/sec. The equation s=(vi)t + at^2 is not found in any textbook unless it is a typo. It should be s=(vi)t + (a/2)t^2. If you knew any calculus, you would understand that. And no, it does not only work for when vi is zero.
From: Don1 on 16 Sep 2005 08:37 odin wrote: Snip< > > The equation s=(vi)t + at^2 is not found in any textbook unless it is a > typo. It should be s=(vi)t + (a/2)t^2. The equation s=(vi)t + (a/2)t^2 is not found in any textbook that I know of: The parentheses are mine! I put them around vi, because vi is a distance l, divided by a unit of time; so that the equation should be s=(l/t)t + (a/2)t^2. The best part is that dividing all terms by t gives s/t=l/t + (a/2)t. Try that in your peace-pipe. Don
From: Randy Poe on 16 Sep 2005 08:50 Don1 wrote: > Randy Poe wrote: > > Don1 wrote: > > > > > > These formulas are just adaptations of those found in most basic > > > physics texts: > > > > "Adapted" in that they were changed from correct to incorrect > > (I missed your error in forgetting to put a "t" after "vi"). > > That wasn't an error Randy: There is no t after vi because vi is a > _ratio_ of length l, to t. The formula in the textbooks is wrong: It > should be s=(v/i)xt + (a/2)t^2: Um... (1) That *is* the formula in textbooks, and (2) You just put the t in after vi, after telling me there isn't one. Though you made another change, from vi for "initial velocity" to v/i for...? Try your version s = vi + at^2/2 for a = 0, constant velocity non-accelerated motion. Do you think s = vi correctly describes the motion? - Randy
From: odin on 16 Sep 2005 09:08 > Um... What I am curious about is that Don1 thinks the textbooks are wrong on this stuff. Yet his equation seems to be different and inconsistent every time he posts a message about it. God only knows how vi became v/i for no good reason in this thread. And God only knows what parenthesis actually mean in his own twisted version of elementary algebra. God only knows why he eschews calculus, vectors, quantum physics, relativity, etc. But in any case, I wonder how he explains that everyone from Ford to NASA uses the appropriate equations as they are described in the textbooks, and they all design devices that actually work. Sometimes he is a throwback to Newton's time, but loaded with errors. Sometimes he is a throwback to ancient Greek times. His math never gets beyond the level of a failing grade seven student. He never gets into the 19th century let alone the last one or this one. Well, if these equations that have been in constant use for over 300 years are wrong, then what the hell does Don1 think is going on? Well, I think I know... Don1 does not believe in what he says. He has a keen interest in annoying others and bringing humiliation on himself. For whatever reasons. Of course, that is just my operating theory. He could also just simply be an idiot.
From: Don1 on 16 Sep 2005 10:48
Don1 wrote: > Randy Poe wrote: > > Don1 wrote: > > > > > > These formulas are just adaptations of those found in most basic > > > physics texts: > > > > "Adapted" in that they were changed from correct to incorrect > > (I missed your error in forgetting to put a "t" after "vi"). > > That wasn't an error Randy: There is no t after vi because vi is a > _ratio_ of length l, to t. The formula in the textbooks is wrong: It > should be s=(v/i)xt + (a/2)t^2: Dividing all terms by t we should get > s/t=l/t + (a/2)t > My mistake Randy: The formula in the textbooks should be s=(l/t)t + (a/2)t^2: So that dividing all terms by t we get s/t=l/t + (a/2)t. Like you told me velocity is a ratio of displacement to time; so that vi=l/t. Don > > However, if you fix that error, then much of what you > > said is not far from correct, although you are misinterpreting > > the meaning of d. > > > Look in your basic physics text: Although they probably use s instead > of d; the equation is given as s=(vi)t + at^2; No wonder it only works > when vi=0'/sec. > > > Don > > - Randy |