# Related to the information contained in $a = v \frac {dv}{ds}$

While studying kinematics I came to the definition of acceleration which is $$a = \frac {dv}{dt}$$. But from this equation we can derive that $$a = v \frac {dv}{ds}$$ which when I evaluate at $$v=0ms^{-1}$$ (here as I know $$v$$ is the instantaneous velocity) gives $$a=0$$ which doesn't seem to be true, for example in the case of projectile at maximum height. So I think that what I'm doing is wrong somewhere but I can't figure out what it is. Can you please help tell what I doing wrong here? Are there any assumptions taken while deriving the given equation for which I'm not accounting for over here?

• $a=\dfrac{dv}{dt}=\dfrac{dv}{ds}\dfrac{ds}{dt}=\dfrac{dv}{ds}\,v$, so if a=0 either v=0 or dv/ds=0
– Eli
Nov 12 '19 at 13:15
• @Elibut what if $v=0$ then is it necessary for $a=0$, if so why?
– user238497
Nov 12 '19 at 13:22
• if v=0 then dv/dt =0 so a=0
– Eli
Nov 12 '19 at 13:25
• if v(t)=0 then dv/dt =0 so a=0
– Eli
Nov 12 '19 at 13:32
• @Ramanujan_π The projectile at maximum height has zero velocity but $lim_{h \to h_{max}} \frac{dv}{dh} = -\infty$. However, $v \frac{dv}{dh} = -g$ holds true at all points of the projectile's trajectory. Nov 12 '19 at 13:33

If you examine the equation $$a = v \frac{dv}{ds}$$ when $$v \rightarrow 0$$, then it turns out that, $$\frac {dv}{ds} \rightarrow -\infty$$. So when $$v$$ is exactly $$0$$, the acceleration takes the indeterminate form of $$a=(0)(-\infty)$$. Let me explain why.

As the velocity tends to $$0$$, the small change in the displacement i.e. $$ds$$ also tends to zero (it gets closer to zero than it was ever before). This can be seen by using the equation, $$v=\frac{ds}{dt} \Rightarrow vdt=ds$$ So clearly when $$v=0 \Rightarrow ds=0$$.

Caution:- I am not saying that $$ds$$ tends to $$0$$ (well, it always tends to zero, at least for our scenario), but here when $$v=0$$, it has an exact value of $$0$$.

Thus now $$\frac{dv}{ds}$$ is no longer a $$\frac{infinitesimal}{infinitesimal}$$ form, but it is a $$\frac{infinitesimal}{0}$$ form which is equal to $$±\infty$$ (the sign of the infinity depends upon the sign of the infinitesimal quantity).

And as explained in the comments, the values of $$\frac{dv}{ds}$$ approaches $$-\infty$$ as the projectile reaches the peak and when the projectile is on its way down, the value of $$\frac{dv}{ds}$$ goes from $$+\infty$$(at the peak) to some finite value.

P.S. Many things might seem too "disgusting" to people who are passionate about mathematics as I have written this answer not using the appropriate mathematical rigour. I have used some vague, yet intuitive vocabulary in this answer. And that's intentional! It is because I want you to get the feel for it rather than getting stuck between evaluating limits.

• Mathematically $ds$ is an infinitesimal but it doesn't tends to $0$ Nov 12 '19 at 14:15
• @Shreyansh Well, it does tend to zero (for all finite velocities) , when $dt$ tends to zero.
– user243267
Nov 12 '19 at 14:16
• Infinitesimal doesn't mean tending to zero Nov 12 '19 at 14:23
• Check out this link:- en.wikipedia.org/wiki/Infinitesimal. See the "Functions tending to zero" section.
– user243267
Nov 12 '19 at 14:36

If you have a parabolic trajectory

$$s = \frac{1}{2} g t^2$$ with velocity $$v = \frac{ds}{dt} = gt$$ then note $$v = \sqrt{2gs}$$ and $$\frac{dv}{ds} = \frac{\sqrt{g}}{\sqrt{2s}}.$$ Note that the function $$1/\sqrt{s}$$ is infinite at $$s = 0$$. So the equation $$a = v \frac{dv}{ds}$$ is true at $$t = 0$$, A.K.A. $$s = 0$$, when $$v = 0$$ because $$\frac{dv}{ds}$$ blows up in just the right way: $$a = v \times \frac{dv}{ds} = \sqrt{2gs} \times \frac{\sqrt{g}}{\sqrt{2s}} = g.$$

• The first equation mentioned is not a equation of parabolic trajectory. Nov 12 '19 at 13:47
• Yes it is, it's the y coordinate of a particle moving with constant acceleration, i.e. $y=-s$. I kept things one dimensional and used $s$ instead of $y$ to make contact with the askers notation. Nov 12 '19 at 14:16