What is v(dv/dx) and how does it relate to acceleration?

[Big-Daddy]

My mechanics syllabus suggests that we can model acceleration as $v(dv/dx)$ but what exactly does this mean? Can you give me (or link me to) some explanations and/or sample problems with worked solutions? In my experience so far we should always be using $dv/dt$ for $a$.

 

[boneh3ad]

Consider what $v$ actually is in one dimension:
$$v = \dfrac{dx}{dt}.$$
So in using the chain rule,
$$v\dfrac{dv}{dx} = \dfrac{dx}{dt}\dfrac{dv}{dx} = \dfrac{dv}{dt} = a.$$
Of course, you could make the same argument with any spatial coordinate. This is intimately related to what is commonly called the material derivative.

 

[Big-Daddy]

Consider what $v$ actually is in one dimension:
$$v = \dfrac{dx}{dt}.$$
So in using the chain rule,
$$v\dfrac{dv}{dx} = \dfrac{dx}{dt}\dfrac{dv}{dx} = \dfrac{dv}{dt} = a.$$
Of course, you could make the same argument with any spatial coordinate. This is intimately related to what is commonly called the material derivative.

Thanks.

What would be the usefulness of something like this? Could you, for example, model motion with air resistance using it? (This is impossible with $dv/dt$ calculus since your balanced expression for acceleration would depend on -kv², your air resistance, but then $v$ depends on acceleration, meaning the model cannot incorporate air resistance.)

 

[olivermsun]

Suppose you have a stream of some fluid and you know the velocity field. You would like to know the acceleration of (and hence the forces on) some bit of fluid as it passes by a certain point. This formulation is useful.

 

[elfmotat]

It's useful when separating differential equations like the following:

$$a=\frac{k}{r^2}$$

So you can integrate:

$$vdv=\frac{k}{r^2}dr$$

This is used, for example, with finding the time it takes for two masses to come together under gravity.