newtonian mechanics - Zero velocity, zero acceleration?

In one dimension, the acceleration of a particle can be written as:

$$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$$

Does this equation imply that if:

$$v = 0$$

Then,

$$\Rightarrow a = 0$$

I can think of several situations where a particle has a non-zero acceleration despite being at instantaneous rest. What's going on here?

Answer

The correct thing to say would be that "if v=0 and dv/dx is finite then a=0".

A simple example, to help illustrate what's going on, is the well known case of constant acceleration "-g" near the earth's surface. In this example, we consider "x" to be the height above the ground, and assume the initial x is zero.

In this case $$ x=-\frac{gt^2}{2}+v_0t $$ $$ v=v_0-gt $$ and $$ a=-g $$ and, clearly, "a" can never be zero, but "v" can be zero... so what gives? Well... solving for t(x) gives $$ t(x)=\frac{1}{g}(v_0\pm\sqrt{v_0^2-2gx}) $$ and $$ v(x)\equiv v(t(x))=\pm\sqrt{v_0^2-2gx} $$ and so $$ \frac{dv}{dx}=\frac{\pm g}{\sqrt{v_0^2-2gx}} $$ ...which is conveniently infinite whenever v is zero.

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Furthermore, I think that a more natural way to think about this issue can be found by considering what we really mean by $$ v(x) $$ and how we go about taking the derivative w.r.t. x.

What we really mean is that, given some functional form for "v" as a function of "t" called "v(t)", and given some functional form for "x" as a function of "t" called "x(t)", and given that "x(t)" can be inverted to find "t(x)", then, as mentioned above $$ v(x)=v(t(x))\;, $$ which is silly physicist notation. It is clearly silly notation because the "v()" on the left-hand side cannot actually have the same form as the "v()" on the right-hand side. That's clear, right? So really let's call it $\tilde v$. I.e., $$ \tilde v(x)=v(t(x)) $$ This function $\tilde v$ is a function of x and the derivative with respect to x is $$ \frac{d\tilde v}{dx}(x)=\frac{dv}{dt}(t(x))\frac{dt}{dx}=\frac{\frac{dv}{dt}(t(x))}{\frac{dx}{dt}}=\frac{a(t(x))}{v(t(x))} $$ I.e., (switching back to silly notation and not writing arguments of functions) $$ \frac{dv}{dx}=\frac{a}{v}\;, $$ so, clearly, for constant a, dv/dx is infinite whenever v=0.