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

This question already has an answer here:

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?

## marked as duplicate by John Rennie, Thomas Fritsch, stafusa, PM 2Ring, Alfred CentauriAug 18 at 11:30

• We know that, $1 + 1 = 2$. But multiplying and dividing by $0$, we get $1 + 1 = 0 * 2 / 0$. According to this equation, $1 + 1 = 0 * (\ldots) = 0$. What is wrong? – knzhou Aug 18 at 6:43
• This is a duplicate of this post. – MannyC Aug 18 at 6:56
• Think in realastic way ,if acceleration is zero ,when velocity is zero ,you will never able to reach the velocity of even 1m/s, – yuvraj singh Aug 18 at 8:22

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.

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.