Acceleration problem

I think Joe's first answer is correct if you interpret the problem as occurring over exactly 1 second and you assume constant acceleration.

Sorry, keep erasing post at last second.

Originally posted by AThousandYoung
I think Joe's first answer is correct if you interpret the problem as occurring over exactly 1 second and you assume constant acceleration.

And I think, or at least I thought that this was the question he intended to ask, but didn't quite know how.

Originally posted by AThousandYoung
Power would be watts, not watt/seconds, but yes.

No, the t and P are not squared.

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/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

Originally posted by sonhouse
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

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.

Pt = K

Pt = mv^2/2

P/t = mv^2/2t^2

P/t = ma^2/2

2P/ta^2 = m

Or, in basic units:

Pt = K

(m d^2/t^3)(t) = (m d^2/2t^2)

(m d^2/t^3)(t) = 1/2 m(at)^2

2(m d^2/t^3)(t)/(at)^2 = m

2P/ta^2 = m

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?

Originally posted by AThousandYoung
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?

ATY - you are the most patient man in the Universe!!

Originally posted by AThousandYoung
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.

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, 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.

Originally posted by sonhouse
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.

Good Luck to your wife, hope all goes well for you both.

Originally posted by wolfgang59
Good Luck to your wife, hope all goes well for you both.

Thanks, I just saw the stitches on her hand, (long story) and they go from the tip to the bottom of her middle finger in a zig zag. Sucks really.

Originally posted by sonhouse
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.

Oh no! I hope your wife heals quickly.

That confusion is why I wrote out the derivation clearly in two different ways.

Originally posted by wolfgang59
ATY - you are the most patient man in the Universe!!

Nah...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 😕

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...

Originally posted by sonhouse
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.

It's written with an a^2 for convenience. Let's see if I can write it in a more intuitive way.

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.

Originally posted by AThousandYoung
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?

It 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.