## Astrophysics one year ago @empty

1. Astrophysics

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

2. Empty

okk I seeee

3. 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)

4. Astrophysics

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

5. 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 $ads=vdv$

6. Astrophysics

|dw:1437972125677:dw|

7. Astrophysics

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

8. Empty

Haha coool

9. Astrophysics

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

10. Astrophysics

mathematically

11. Astrophysics

mhmm

12. 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?

13. Astrophysics

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

14. Astrophysics

@Empty

15. 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

16. 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

17. Empty

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

18. 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!