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?
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1$\begingroup$ $ 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 $\endgroup$– EliCommented Nov 12, 2019 at 13:15
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$\begingroup$ @Elibut what if $v=0$ then is it necessary for $a=0$, if so why? $\endgroup$– user238497Commented Nov 12, 2019 at 13:22
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$\begingroup$ if v=0 then dv/dt =0 so a=0 $\endgroup$– EliCommented Nov 12, 2019 at 13:25
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$\begingroup$ if v(t)=0 then dv/dt =0 so a=0 $\endgroup$– EliCommented Nov 12, 2019 at 13:32
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1$\begingroup$ @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. $\endgroup$– Ajay MohanCommented Nov 12, 2019 at 13:33
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2 Answers
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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.
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$\begingroup$ Mathematically $ds$ is an infinitesimal but it doesn't tends to $0$ $\endgroup$ Commented Nov 12, 2019 at 14:15
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$\begingroup$ @Shreyansh Well, it does tend to zero (for all finite velocities) , when $dt$ tends to zero. $\endgroup$– user243267Commented Nov 12, 2019 at 14:16
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$\begingroup$ Infinitesimal doesn't mean tending to zero $\endgroup$ Commented Nov 12, 2019 at 14:23
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$\begingroup$ Check out this link:- en.wikipedia.org/wiki/Infinitesimal. See the "Functions tending to zero" section. $\endgroup$– user243267Commented Nov 12, 2019 at 14:36
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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: $$ a = v \times \frac{dv}{ds} = \sqrt{2gs} \times \frac{\sqrt{g}}{\sqrt{2s}} = g. $$
answered Nov 12, 2019 at 13:38
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$\begingroup$ The first equation mentioned is not a equation of parabolic trajectory. $\endgroup$ Commented Nov 12, 2019 at 13:47
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$\begingroup$ 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. $\endgroup$ Commented Nov 12, 2019 at 14:16