# Terminology for time derivative of speed (not velocity)

Is there any standard terminology for the derivative of the magnitude of velocity with respect to time (suitable for use in first-year Calculus)? The word ‘acceleration’, in its technical sense, is exactly what I am not looking for; it is the derivative of the velocity itself, but I want the derivative of its magnitude, the speed.

This is a useful concept, because it matches the colloquial sense of the word ‘acceleration’; if this quantity is positive, then an object is speeding up (accelerating), if it is negative, then it is slowing down (decelerating), and if it is zero, then it's doing neither (although it might be changing direction). As $$v$$ is sometimes used for the magnitude of the velocity vector $$\boldsymbol v$$, so $$a$$ is sometimes used for the quantity that I'm looking for, but notice that $$a = \mathrm d v / \mathrm d t$$ is not the magnitude of the acceleration vector $$\boldsymbol a = \mathrm d \boldsymbol v / \mathrm d t$$. (After all, $$a$$ can be negative.)

In the $$1$$-dimensional case, $$\boldsymbol a = \pm a$$ (although you might not want to use those symbols in this case), but the $$\pm$$ is given by the sign of $$\boldsymbol v$$ rather than by the sign of $$\boldsymbol a$$. In more dimensions, you can write $$\boldsymbol a = a \boldsymbol T + \kappa v ^ 2 \boldsymbol N = a \boldsymbol T + \omega v \boldsymbol N$$, where $$\boldsymbol T = \boldsymbol v / v$$ is a unit vector in the direction of motion (so the $$a \boldsymbol T$$ term is analogous to the $$\pm a$$ in $$1$$ dimension), $$\boldsymbol N$$ is a unit vector in the direction of curvature, $$\kappa$$ is amount of curvature, and $$\omega = \kappa v$$ is angular speed. (In $$3$$ dimensions, angular velocity relative to the centre of the osculating circle is $$\boldsymbol \omega = \omega \boldsymbol B$$, where $$\boldsymbol B = \boldsymbol T \times \boldsymbol N$$ is the unit binormal vector, but angular speed makes sense in any number of dimensions.) So this is certainly a useful concept for analysing acceleration, in particular breaking acceleration down into change of speed and change of direction.

I have called this ‘colloquial acceleration’, but I haven't seen that used anywhere else. I've also called it ‘scalar acceleration’, and so have many other people, but that's not a good term if you're working in only $$1$$ dimension (where everything is a scalar). I've also seen ‘tangential component of [the] acceleration’ (with ‘normal component of [the] acceleration’ for $$\kappa v ^ 2 = \omega v$$), which is not so much a term for it as a definition of it, but it also doesn't work very well in $$1$$ dimension.

So, is there a standard term for this? If people think that either of the vector-related terms is sufficiently standard that I could say with a straight face to my one-variable Calculus students ‹The term for the derivative of speed with respect to time is is ‘scalar acceleration’; next year, you will learn what ‘scalar’ means, but for now it is just technical jargon.›, then I will use it. Or if there is another standard term for it that I haven't heard of, then I will use that. If not, then I'll stick to ‘colloquial acceleration’, even though it is clearly not standard, since it's easy to explain.

You answer your question in your question. You are wanting the tangential component of the acceleration. That's the name of it. That is the component of the acceleration that changes the object's speed.

The other component you mention is referred to as the radial component, or the radial acceleration. It's responsible for changing the direction of the velocity.

• To clarify, you're saying that ‘tangential component of the acceleration’ is the standard name, and even when talking to beginning Calculus students working with motion in 1 dimension, it would be appropriate to say ‹The derivative of the speed with respect to time is called the tangential component of the acceleration; next year, when you study vectors, you will learn what ‘tangential component’ means, but for now it is just technical jargon.›? Commented Aug 6, 2019 at 19:15
• @TobyBartels Yes, if that is what you want to do. That is what changes the speed. It doesn't fix the sign issue though. The tangential acceleration could be constant with the speed decreasing and then increasing. Commented Aug 7, 2019 at 2:12
• For my purposes, the sign issue is vital. If the speed is decreasing and then increasing, then its derivative is negative and then positive, so not constant. Commented Aug 7, 2019 at 20:23
• @TobyBartels Yeah you are going to have an issue there. For example, if something is moving to the right and I apply a force to the left the speed will decrease and then increase (the object will stop and then start moving to the left). So your "speed derivative" will change directions, but nothing physically has changed. There really isn't any way around it because it doesn't generalize well. You can't say a certain force causes a certain speed derivative. It depends on the motion of the object. Commented Aug 7, 2019 at 22:34

I want the time derivative of speed. This is a useful concept, because it matches the colloquial sense of the word ‘acceleration’; if this quantity is positive then an object is speeding up (accelerating), if it is negative then the object is slowing down (decelerating), and if it is zero then the object is doing neither (although it might be changing direction).

I think that the rate of change of speed is the most succinct descriptor of the time derivative of speed. Alternatively: the tangential component of acceleration, i.e., the component of acceleration in the direction of motion. And, since retardation is defined as the rate of decrease of speed, it can also be characterised as the additive inverse of retardation.

By the way, the particle with position function $$-e^{t}$$ is speeding up at a negative acceleration (the minus sign referring to spatial direction), so I'd hesitate to describe it as "accelerating".

Finally, the position function $$\cos(t)\mathbf{i}+\sin(t)\mathbf{j}+t\mathbf{k}$$ dispels the notion that the time derivative of speed is the magnitude of acceleration: the derivative of its speed is $$0,$$ whereas the magnitude of its acceleration is $$1.$$

• It's not a perfect answer (since it's the opposite of what I want), but ‘retardation’ seems like a good word to know, thanks. Commented May 26 at 14:01