Posers and Puzzles
18 Jan 10
- Joined
- 10 Dec 06
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19 Jan 10
Originally posted by AThousandYoungAnd I think, or at least I thought that this was the question he intended to ask, but didn't quite know how.
I think Joe's first answer is correct if you interpret the problem as occurring over exactly 1 second and you assume constant acceleration.
20 Jan 10
3 edits
Originally posted by AThousandYoungso is your formula rigorous, would stand as written in a calculator or computer with say mathematica? Not sure because this is not a rigorous math notation, we can't do it right here. Can you give it to me with correct parentheses?
Power would be watts, not watt/seconds, but yes.
No, the t and P are not squared.
As you have it: m=P/ta^2 the thing I am not sure about here is the ^2. Is that to be applied only to a or should it be m=(P/ta)^2?
It seems kind of weird to take 'a' which already has a ^2 function and ^2 function it again. 'a' being feet/second^2
20 Jan 10
5 edits
Originally posted by sonhousePEMDAS!
so is your formula rigorous, would stand as written in a calculator or computer with say mathematica? Not sure because this is not a rigorous math notation, we can't do it right here. Can you give it to me with correct parentheses?
As you have it: m=P/ta^2 the thing I am not sure about here is the ^2. Is that to be applied only to a or should it be m=(P ...[text shortened]... take 'a' which already has a ^2 function and ^2 function it again. 'a' being feet/second^2
Exponents before Multiplication and Division!
m = 2P/[t(a^2)]
if that helps, but those parentheses are redundant. Exponents are evaluated before multiplication or division. I put the "correct" parentheses - none.
Yes, my formula is rigorous. The single power of t comes from the Pt = K.
20 Jan 10
Originally posted by AThousandYoungATY - you are the most patient man in the Universe!!
Really, it seems like the definition of kinetic energy should make this clear:
K = 1/2 mv^2
Energy is proportional to the square of the velocity. If you add energy at a steady rate, why would you expect velocity to increase steadily as well - that is, proportionally?
20 Jan 10
Originally posted by AThousandYoungOk, just wanted to be sure what you wrote and what I read is the same thing! The Parentheses are redundant but it helps to clarify the formula when you can't use mathematica or lab view or some such. I'm sure you have been messed up by small transitions of formula's that completely changes the outcome.
PEMDAS!
Exponents before Multiplication and Division!
m = 2P/[t(a^2)]
if that helps, but those parentheses are redundant. Exponents are evaluated before multiplication or division. I put the "correct" parentheses - none.
Yes, my formula is rigorous. The single power of t comes from the Pt = K.
I have not done a close look at that stuff yet, will do in the next few days, have an injured wife to attend to, and it is going to get worse before it gets better, she has an upcoming operation to replace both knee's at the same time so I don't know how much longer I can come back to RHP for the next three or four months.
20 Jan 10
Originally posted by sonhouseGood Luck to your wife, hope all goes well for you both.
Ok, just wanted to be sure what you wrote and what I read is the same thing! The Parentheses are redundant but it helps to clarify the formula when you can't use mathematica or lab view or some such. I'm sure you have been messed up by small transitions of formula's that completely changes the outcome.
I have not done a close look at that stuff yet, wil ...[text shortened]... e so I don't know how much longer I can come back to RHP for the next three or four months.
21 Jan 10
Originally posted by sonhouseOh no! I hope your wife heals quickly.
Ok, just wanted to be sure what you wrote and what I read is the same thing! The Parentheses are redundant but it helps to clarify the formula when you can't use mathematica or lab view or some such. I'm sure you have been messed up by small transitions of formula's that completely changes the outcome.
I have not done a close look at that stuff yet, wil ...[text shortened]... e so I don't know how much longer I can come back to RHP for the next three or four months.
That confusion is why I wrote out the derivation clearly in two different ways.
21 Jan 10
2 edits
Originally posted by wolfgang59Nah...I'm writing my thoughts as they arrive. When I saw this problem I made the same assumption sonhouse did. Then I came up with extra unknown variables when I did the math 😕
ATY - you are the most patient man in the Universe!!
Besides I know sonhouse is extremely intelligent, especially with scientific matters, and is a very nice guy to boot. We're a couple of the regulars in this forum.
Then again I am halfway done with my Master's in Education...
21 Jan 10
1 edit
Originally posted by sonhouseIt's written with an a^2 for convenience. Let's see if I can write it in a more intuitive way.
Ok, just wanted to be sure what you wrote and what I read is the same thing! The Parentheses are redundant but it helps to clarify the formula when you can't use mathematica or lab view or some such. I'm sure you have been messed up by small transitions of formula's that completely changes the outcome.
I have not done a close look at that stuff yet, wil e so I don't know how much longer I can come back to RHP for the next three or four months.
Pt = K
Pt = mv^2/2
2Pt/v^2 = m
That's equivalent, and maybe more intuitive. The problem is we know the acceleration - it was given - and not the velocity, so the acceleration version is more useful.
- Joined
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21 Jan 10
Originally posted by AThousandYoungIt does, when a bullet is fired every thing else in the system moves. Fire a bullet to the right, the universe moves to the left. Its consevation of momentum.
That has always bugged me.
It seems like it should take the same energy to fire a bullet as for the bullet to stay still and everything else move. But it doesn't...does it?