# How do you correctly apply $a=v\cdot\text dv/\text ds$? [closed]

I know that $$a=v\cdot\text dv/\text ds$$ can be deduced by simple algebra and calculus and is correct. But once I was analyzing the motion of a ball projected straight up in free gravity. If I apply $$a=v\cdot\text dv/\text ds$$ at top most point it gives $$a=0$$, while it should be $$g$$.

Can someone please explain where I am going wrong?

## closed as off-topic by ahemmetter, Jon Custer, JMac, Kyle Kanos, stafusaJul 30 at 14:41

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• No one can say where you went wrong. You didn't show your work. – Aaron Stevens Jul 29 at 12:24
• OP described the problem pretty clearly. – user37222 Jul 29 at 12:39
• @user37222 I didn't say the problem was unclear. I said they didn't show their work so no one can tell them why they are getting a wrong answer. There can be speculation, but nothing certain. – Aaron Stevens Jul 29 at 12:45

There's nothing wrong with $$a=v \frac{dv}{ds}$$. At the highest point, $$v=0$$, which makes $$dv/ds$$ undefined. If you take the appropriate limit, you get $$a=-g$$ again. Let me elaborate.

Start with the standard uniformly accelerated linear motion equations:

$$v=v_0 + at$$ $$s = s_0 + v_0 t + \frac{1}{2}at^2$$

Your formula for $$a$$ then gives

$$a = v \frac{dv}{ds} = v \times \frac{dv/dt}{ds/dt} = v \times\frac{a}{v}$$

If you take $$\lim_{v \to 0}$$, this gives $$a$$.

Otherwise, you get an undefined number.

• Why is $dv/ds$ undefined when $v=0$? One is a value at a point in time & the other a differential relation. Additionally, $v$ is in the numerator, so what is wrong with 0 being there? Also, the second half of this answer is a useless non sequitur. I don't know how this managed any upvotes. – Kyle Kanos Jul 30 at 11:55
• @KyleKanos For constant acceleration motion in 1D where the acceleration is opposite of the velocity, $dv/ds$ is in fact undefined when $v$ changes sign. At this point the position is not changing, so it's a dividing by $0$ situation. Work it out for $v^2=v_0^2+2as$. You will find the derivative to be undefined for when $s=-v_0^2/2a$ – Aaron Stevens Jul 30 at 15:38
• @AaronStevens yes, $ds=0$ is an is an issue, but that surely is not what is written. – Kyle Kanos Jul 30 at 16:31
• @KyleKanos I was answering your question as to how $dv/ds$ is undefined. If you were to graph $v(s)$ for this scenario you would see that when $v=0$ the tangent of the curve is completely vertical. It is an undefined derivative. – Aaron Stevens Jul 30 at 16:32
• @Aaron you do realize that I do know the answer, right? That the point I'm making is that OPs description of the reason is bad & needs updating, right? – Kyle Kanos Jul 30 at 17:50

If the initial velocity is $$v_0$$, then we have $$v(s) = \sqrt{v_0^2-2gs}$$.

At $$s_0 = \frac{v_0^2}{2g}$$ we have $$v(s_0) = 0$$ but the function $$v(s) = \sqrt{v_0^2-2gs}$$ is not differentiable at $$s_0$$. Therefore the relation $$a(s) = v(s)\cdot \frac{dv(s)}{ds}$$ is actually not true at $$s_0$$, because $$\frac{dv(s)}{ds}\Big|_{s=s_0}$$ does not exist.