## Question about ẍ = v.dv/dx .

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a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Question about ẍ = v.dv/dx .

We know from basic kinematics that for a particle moving in a straight line, its acceleration ẍ satisfies ẍ = v.dv/dx, where v is the velocity and x is the displacement.

If we naïvely plug in v = 0 to this equation, it makes it seem as though the acceleration is 0 whenever the velocity is 0, but we know this is not true (as a basic counterexample, Simple Harmonic Motion clearly has non-zero velocity at the point where acceleration is 0, which is at the centre of motion).

Why does this formula not work if we plug in v = 0 like this? I would like confirmation, but I think it is basically because dv/dx is actually undefined at v = 0 and since in the derivation of this formula, we essentially do:

ẍ = dv/dt

= 1/v * v * dv/dt (so v can't be 0 here)

= dt/dx * v * dv/dt (using 1/v = 1/(dx/dt) = dt/dx, which assumes v is non-zero)

= v.dv/dx (chain rule).
joe
Posts: 755
Joined: Fri Aug 29, 2003 11:57 am
Location: Sydney

### Re: Question about ẍ = v.dv/dx .

I agree. when v goes to zero, dx goes to zero.

Happy new year
Joe
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Question about ẍ = v.dv/dx .

joe wrote:I agree. when v goes to zero, dx goes to zero.

Happy new year
Joe
Thanks, Joe. Happy New Year to you too.