# Is the product of P actually wug and what about

by Probie1
Tags: product
 P: 38 P = mv so do this mean that the product of v is μg and the product of m is weight? So it could be written P = wμg How is this formula derived Vf = √(Vi^2 + (2ad))
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P: 26,148
Hi Probie1!
 Quote by Probie1 P = mv so do this mean that the product of v is μg and the product of m is weight? So it could be written P = wμg
Sorry, I've no idea what you're talking about
what is the context (and what do you mean by "product")?
 P: 38 Does product not mean... umm the make up... it is part of or makes up? I guess the context of all this is I am trying to undertand how formula's come about. P = mv so do this mean that the product of v is μg and the product of m is weight? So it could be written P = wμg This is another question. How is this formula derived Vf = √(Vi^2 + (2ad))
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,502 Is the product of P actually wug and what about Mathematically "product" means the result of multiplying numbers. It simply doesn't make sense to talk about the "product" of a single number as in "product of v is μg" or "the product of m is weight". Perhaps you mean it the other way- weight is the product of mg. That is "mass times the acceleration due to gravity of an object is the force on that object due to gravity"- by definition its "weight". I'm not sure what you could mean by "v is the product μg", if that is what you intend, because you have not told us what μ is and it is not a standard symbol. Sometimes μ is used for the "coefficient of drag" but that doesn't make sense here. Assuming g is the acceleration due to gravity and v is velocity, their standard meanings, since v would have units of "meters per second" and g "meters per seconds squared", μ would have to have units of "seconds"- it would have to be a "time". Is that correct?
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 Quote by Probie1 This is another question. How is this formula derived Vf = √(Vi^2 + (2ad))
What other equations of motion do you know?
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Hi Probie1!

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 Quote by Probie1 How is this formula derived Vf = √(Vi^2 + (2ad))
This is one of the standard equations for constant acceleration.

Then, integrating, v = at + vi.

And integrating again, d = 1/2 at2 + vit.
Can you finish the proof?
 Quote by Probie1 Does product not mean... umm the make up... it is part of or makes up?
As HallsofIvy says, no.

What did you mean by P m v m and g ?
P: 38
 Can you finish the proof? d = 1/2 at2 + vit.

D= at + vi2

Alright... stop laughing.

 What did you mean by P m v m and g ?
P = momentum
m=mass
v=velocity
g = gravity
μ = friction
w= weight

I thought that if P=mv then v = a = μg but then I remembered where I left my brain because a = change in velocity over a change in time. So it can't possibly be the way I was thinking. So just forget I was so stupid to write that down.

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