If acceleration $a = v\cdot\frac{\mathrm dv}{\mathrm dx}$ , then why isn't acceleration always zero when velocity is zero?

Question

We know that,

$$\begin{align} a&=\frac{\mathrm dv}{\mathrm dt}\\[3pt] &=\frac{\mathrm dv}{\mathrm dx}\cdot\frac{\mathrm dx}{\mathrm dt}\\[3pt] &=v\cdot~\frac{\mathrm dv}{\mathrm dx} \end{align}$$

According to this equation, whenever velocity is $0$, acceleration becomes $0$ as well.

However, we know that this is not always the case.

For example, when we throw a ball upwards, at its maximum height, its velocity becomes $0$ but the acceleration is non-zero ($g$ downwards).

Can someone please help me understand what is wrong in my approach?

Answer 1

It is perhaps worth considering your equation in terms of finite changes rather than infinitesimal changes.

If $a = \dfrac {\Delta v}{\Delta t}$ then for a given interval of time $\Delta t$ there is a fixed change in the velocity $\Delta v$.

$a = \dfrac{\Delta v}{\Delta x}\dfrac{\Delta x}{\Delta t}$

If the second (velocity) term on the right hand side of the equation for acceleration is getting smaller and smaller then $\Delta x$ must be must be getting smaller and smaller.

If $\Delta x$ is getting smaller and smaller then the first term is getting bigger and bigger.

Since the rates of increase and decrease for the two terms, which are both dependent on $\Delta x$, are the same then the product of the two terms stays constant.

Note that the first term blows up if you try and make $\Delta x$ equal to zero.

Answer 2

Calculus doesn’t work that way. You can’t regroup $(dv/dv)(dv/dt)$ as you’re trying to.

The basic parts of $dv/dt$ are $v$, the velocity, and $d/dt$, an operator that forms a derivative.