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**[SOLVED] A very simple question about velocity and acceleration**

## Homework Statement

I should be way over this by now (I took elementary mechanics a year ago), but occasionally I find out there's unacceptably much I don't understand about very elementary physics (perhaps is it a sign to be taken seriosly about the future of my physics career...). And this question in particular is ridiculously easy, I guess.

In Fowles's and Cassiday's

*Analytical mechanics*it says:

It goes on to say thatIf the force is independent of velocity or time, then the differential equation for rectilinear motion is simply:

[tex]F(x)=m\ddot{x}[/tex]

[tex]\ddot{x}=\frac{d\dot{x}}{dt}=\frac{dx}{dt}\frac{d\dot{x}}{dx}=v\frac{dv}{dx}[/tex]

Now what I don't understand is that if F is a function of x, then acceleration (a) should be a function of x also. However

[tex]\ddot{x}=\frac{d^2x}{dt^2}=a=\frac{dx}{dt}\frac{d\dot{x}}{dx}=v\frac{dv}{dx} \not= va[/tex]

although if F=F(x) and a=a(x) then shouldn't

[tex]\frac{dv}{dx}=a[/tex]?

On the other hand the book states that

[tex]F(x)=mv\frac{dv}{dx}[/tex]

and if I divide by m I get:

[tex]\frac{F}{m}=a=v\frac{dv}{dx}[/tex]

So dv/dx is not a in this case but dv/dx

**times v**. But if the force is independent of time, then how can velocity still be dx/

**dt**? How can the position derivated with respect to time give out velocity in full, if there's an acceleration affected by the position and therefore a velocity affected by the position?

So I guess I'm having trouble understanding even the initial quotation. I know how it is, but I don't

*understand*why it is.