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- Astrophysics

Say |dw:1437971434694:dw| so for velocity we get, \[v = \frac{ ds }{ dt }\] for a particle moving through the displacement delta s yeee?

- Empty

okk I seeee

- Astrophysics

So for acceleration then, say |dw:1437971693486:dw| we get \[a = \frac{ dv }{ dt }\] (I made pewww pewww noises when I wrote down velocity, I can never be a serious physicist)

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- Astrophysics

So doing some substitution we get \[a = \frac{ d^2s }{ dt^2 }\]

- Astrophysics

So now say we have this diagram |dw:1437971855435:dw| now it's decelerating. We see here \[\Delta v = v' - v\] will be negative so a will also be negative but it says we will get a differential relation involving displacement, velocity, and acceleration along the path may be obtained by eliminating the time differential dt between \[v = \frac{ ds }{ dt }~~~\text{and}~~~a = \frac{ dv }{ dt }\] giving us \[ads=vdv\]

- Astrophysics

|dw:1437972125677:dw|

- Astrophysics

I guess I could've skipped all that and just went to the equations xD

- Empty

Haha coool

- Astrophysics

Yeah, I just don't really see the ads=vdv part

- Astrophysics

mathematically

- Astrophysics

mhmm

- Empty

Hmmm I guess in one way we could write it as:
\(ads = vdv\) we can divide to both sides like this:
\[\frac{a}{v} = \frac{dv}{ds}\]
\[\frac{\frac{dv}{dt}}{\frac{ds}{dt}} = \frac{dv}{ds}\]
Like that?

- Astrophysics

No I mean why is it ads = vdv, why is it set up like that? What does it mean

- Astrophysics

- Astrophysics

I tried getting that expression without going backwards, and I can't lol, like I started with just \[a = \frac{ dv }{ dt }\] and ended up with \[v = \frac{ dv }{ ds }\] but I don't know if this is even right..lol

- Empty

Hmmmm I am not entirely sure what your question is, like "why is it this way" or "how does it end up this way" and I think I can only answer the second question haha

- Empty

Hehe we can shift the numerators on "dt" to get to the other side of that middle equals sign there kinda haha
$$a ds = \frac{dv}{dt}ds = dv\frac{ds}{dt} = dv v$$

- Astrophysics

Lol well thanks xD, I see what you're doing, but only I can answer my own question I think because it's probably just something trivial...thanks mang!

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