Exploring the quantity dv/ds

Arnold Arons, in teaching introductory physics, asks us to consider the quantity dv/ds as an alternative possible concept for acceleration. Arons claims that Galileo considered both dv/dt and dv/ds. He also suggests that Galileo gave some less-than-compelling arguments against dv/ds, favoring dv/dt because of his hunch that it better described free-fall. Arons invites us to explore the meaning of dv/ds as a concept for changing speed, so here are some of my explorations.

Constant Alternative Acceleration

I started by considering the case of 1D constant alternative acceleration, and worked to rewrite it in terms of the more familiar time-parametrized trajectories.

First let’s just consider what a constant dv/ds means qualitatively. This means that you gain the same amount of speed in an equal amount of distance covered. This makes sense from the units, which are (m/s)/m or  1/s.

Mathematically, we can approach this the following way:

dv/ds = constant (we’ll just call it b), from which is follows that

dv = b ds

From which, we’ll integrate, and for now assume some positive initial velocity vo, giving us

v = b (s-so) + vo

Let’s just set so = 0. Now, we want to integrate one more time to get us an expression for position vs time, so we’ll rewrite v as ds/dt

ds/dt = bs + vo

And then we’ll separate variables and integrate again, from s = 0 to s = s’, and t =0 to t=t’ (and of course rewrite t’ as t and s’ as s)

ds / (bs + vo) = dt

1/b log(bs+vo) – 1/b log(vo) = t

Then it’s just a rearrange game:

log [ bs/vo +1] = bt

b/vo s + 1 = Exp[bt]

s * b/ vo = Exp[bt] -1

s = vo/b (Exp[bt]-1)

Taking a derivative to get v as an expression in terms of t, we get  v = vo Exp(bt), which shows that for positive values of b the speed exponentially grows a, and for negative values of b, the speed exponentially decays. What is interesting is that constant alternative acceleration has no turn around. It’s either run-a-away speeding up or taking forever to slow down. Well, of course is b=0, then velocity is constant.

There is another interesting feature of constant alternative acceleration. Consider the case that vo = 0. For that case we’ll can back up in our derivation, and not assume s0 = 0.

ds/dt = b(s-so)  + 0

This says that if you are located where you start, then you have no velocity. And then, the only way to can get somewhere not where you started, is to have some velocity. So I think this means that if you start with a velocity = 0, then no matter how much alternative acceleration you have, you will never get anywhere. This is strange: Any finite alternative acceleration cannot get an object moving.

Constant Normal Acceleration viewed from the Alternative Acceleration

Now, what if we look at it backwards, how do we describe normal constant acceleration in terms of our alternative constant acceleration. Let’s do that.

dv/dt = a

(dv/ds) (ds/dt) = a

(dv/ds) * v = a

dv/ds = a/v

This says that the “alternative acceleration”, which is obviously not constant, decreases as velocity gets big. This makes sense, because as you move fast, you cover a lot of distance in a lot of time, so you don’t change speed very much over that distance. What’s interesting is that we now have a “zero” in the denominator to contend with, where the alternative acceleration is not-defined where the velocity is zero. But we know this must be the case (from my previous exploration):  If v=0, any finite amount of acceleration cannot get an object moving.

Another interesting feature is that the alternative acceleration changes sign as velocity changes sign. For a tossed ball, on the way up, the alternative acceleration I think would be negative and approaching infinity, then it’s undefined, then it becomes positive moving away from positive infinity approaching zero asymptotically. Oddly enough, I think this connects somehow to the puzzle described here.

A physical situation where alternative acceleration is simple

One more case I’d like to consider, because it’s simple is the case of linear drag force. From Newton’s Second Law, we get.

m dv/dt = -cv

using the same trick above, we write this as

v dv/ds = -c/m v

As long as v isn’t equal to zero, we get

dv/ds = -c/m

Which shows that linear drag has constant alternative acceleration… which makes sense from my exploration above, where we round that for negative values of alternative acceleration, the speed approaches zero asymptotically, taking forever to slow down completely.

Other commentary, thoughts, and explorations…

dv/ds as a definition of acceleration is inertial-frame dependent, which tells us something important about how it can relate to a Newtonian perspective.

It’s interesting to think about what the next derivative should be to describe how alternative acceleration changes: d(dv/ds)/dv?  or d(dv/ds)/ds ? or d(dv/ds)/dt or something else?

dv/ds is, of course, directly related to slope of the phase space. Plotting the phase space for various motions helps you get a feel for why dv/ds behaves as it does and especially why it approaches infinity when it does.

Some related and interesting alternative definitions to consider are the quantities  d|v|/ds  and d|v|/d|s| . One way of approaching thinking about these is through reflections of the phase space… Other odd ones to consider are both d(v²)/dt  and d(v²)/ds, wherein we are trying to describe the motion of objects that lose kinetic energy either at a constant time-rate or a constant position-rate. The constant position-rate one is easy… as we’ve actually already explored it.


4 thoughts on “Exploring the quantity dv/ds

Add yours

    1. They are called differentials. It is a common notation in calculus. The “d” you can think of denoting “a little bit of”, or a “small change to”. And then v is velocity, t is time, and s is position. So dt might be though if as a little bit of time or a small change in time; ds a small change in position, and dv a small change in speed. The equations then relate how small changes in, say, time relate to how the object changes in position and velocity.

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