# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ahemmetter, Jon Custer, JMac, Kyle Kanos, stafusa
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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.

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

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.