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Relation between acceleration and velocity 
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#1
Mar1313, 04:40 AM

P: 125

Hi all,
so my question is can i carryout normal algebraic operations on derivatives, for example: v=ds/dt and a=dv/dt then eliminating dt a=(dv/ds) *v then, a *ds= v*dv is that how you derive the relationship between acceleration, velocity and displacement? 


#2
Mar1313, 05:11 AM

Mentor
P: 18,293

The point is that ##ds## and ##dv##are notations that are undefined. So writing ##a ds = v dv## is undefined. In particular, the notation ##\frac{dv}{dt}## is not a fraction. It is something that merely behaves like a fraction.
That said, there are ways to define ##ds## and ##dv##. But this is usually not done in a calculus course, so I won't say anything about that. In short: unless you ever define ##ds##, what you wrote down is meaningless. 


#3
Mar1313, 05:26 AM

P: 125

So then could i define it as a differential change in distance?



#4
Mar1313, 05:28 AM

Mentor
P: 18,293

Relation between acceleration and velocity



#5
Mar1313, 05:31 AM

P: 125

Hmmm, why then can a physics textbook throw that equation out at you, and then expect you to integrate it to find one of those kinematic equations describing motion with constant acceleration assumed?



#6
Mar1313, 05:35 AM

Mentor
P: 18,293

What they do actually does work. And you can explain it physically. But it's mathematically incorrect (unless you actually define ##ds##). I'm going to move this to the physics section to let you get a physicists' perspective on this. 


#7
Mar1313, 06:17 PM

PF Gold
P: 953

It works, and it is, as you suggest, one way of deriving the relationship between start velocity, finish velocity, displacement and acceleration for motion with a constant acceleration. I think all you need do to make it mathematically respectable is (a) to regard your first step, not as eliminating dt, but as applying the chain rule: a = dv/dt = (dv/ds)(ds/dt) = (dv/ds)v, (b) to put integral signs in front of both sides of your last equation.



#8
Mar1413, 06:51 AM

Sci Advisor
P: 2,109

Here are those three ways of interpreting your result [itex]a \cdot ds= v \cdot dv[/itex]:



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