- #1

- 1,195

- 512

Part 1

1. Say we restrict ourselves to one dimension and define a spatial coordinate, x. Then we square it, so now we have x^2.

2. Now we make an equation of it, x^2 = x^2.

3. Now I want to divide each side by 1/t^2, yielding (x^2)/(t^2) = (x/t)(x/t) = v^2, where v=velocity.

Does that look ok so far, can I do that operation?

4. Now I want to redo step 3 in a different manner. Instead of what I wrote above, I want to do the following: divide each side by 1/t^2, yielding (x^2)/(t^2) = (x/t^2) (x) = (a)(x), where a=acceleration and x=spatial displacement along the x coordinate.

My question is, can I take my choice as to how I can split up the time variables on the RHS of the equation in both 3 and 4 given what’s on the left?

5. What about the differential form of 3 and 4? Can I write the RHS version of 3 as (dx/dt)(dx/dt), and version 4 as (d^2x/dt^2)(dx)?

Part 2

I’m interested in the math aspect of the equation but it was the work energy theorem that got me thinking about it. Through a few mathematical tricks I can comfortably follow, the work energy theorem gives you a kinetic energy term, ½ mv^2 from the work term, or, Force times distance (m)(a)(x) = ½(m)(v^2), where m is mass, v is velocity, a is acceleration, and x is spatial displacement.

Now, if we add a mass term to

*my*equation 3 above in part 1, you get mv^2, and if you add that same mass term to equation 4 you get (m)(a)(x), or the work term.

So, if it isn’t obvious already, my conundrum here is that, from my mathematical tinkering in numbers 1-4 above in part 1, I came up with an equivalence (m)(a)(x) = mv^2, but the work-energy theorem tells us that (m)(a)(x) = 1/2mv^2. Where did that extra ½ term come from?