Relation between acceleration and velocity

Clever_name

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?

micromass

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.

Clever_name

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

micromass

What do you mean with that?? Can you write it down formally?

Clever_name

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?

micromass

Physics textbooks aren't always mathematically correct.

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.

Philip Wood

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.

stevendaryl

Derivatives are not ordinary fractions, they are limits. But it is quite common in derivations to treat differentials as if they were small numbers, and manipulate derivatives as if they were ratios of small numbers. But it's a heuristic--it's a way to get a hint as to the form of the answer, but it's not rigorous. Usually, differentials (quantities like ds) are just used as intermediate steps in a nonrigorous derivation. There are three ways that a differential can appear in the final result:

1. As a part of a derivative.
2. As an integrand to an integral.
3. As a finite difference equation.

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

1. As a part of a derivative: [itex]a \cdot \dfrac{ds}{dt} = v \cdot \dfrac{dv}{dt}[/itex]
2. As an integrand to an integral: [itex]\int a \cdot ds = \int v \cdot dv[/itex]. If we multiply the left side by the mass [itex]m[/itex], then it gives the work done by the force acting on the object: [itex]W = \int F \cdot ds = \int m a \cdot ds[/itex]. On the right side, if we multiply by [itex]m[/itex], then it gives the change in kinetic energy:
   [itex] m \int v \cdot dv = \int d(\frac{1}{2} m v^2)[/itex]. So this says that the work done by the forces acting on an object is equal to the change in the kinetic energy of the object.
3. As a finite difference equation. It's approximately true that [itex]a \delta s = v \delta v[/itex], where [itex]\delta s[/itex] is the change in position during a small time period [itex]\delta t[/itex], and [itex]\delta v[/itex] is the change in velocity during that time period.