What is the Change in Acceleration Called?

I added my own input to a recent physics question summitted by another board member. As you can see it has been a while since I’ve taken any physics course.

But this question got me thinking. About Newton’s laws of motion.

Velocity is defined as the change in distance, or simply distance divided by time: D/T. When you look for the change in velocity in relation to time, you get acceleration which is simply V/T–also variously defined as D/T/T and D/T^2. (Remember in the site I refer to, the acceleration due to gravity is given as 9.8m/sec^2.)

Now, my question. What is the change in acceleration in relation to time? What is it called–and how is it used?

I recall hearing somewhere that Calculus plays a role in this all. That one equation is a derivative of another, which is itself a derivative of another and so forth. Does this have anything to do with my answer?

TTFN:D

You could extend this again to wonder what is the name for the change of (rate of change of acceleration) and then the change of(change of (rate of change of acceleration)); it may be that nobody has ever bothered inventing a single-word term for that.

While Mangetout has made the point that the rate of change of acceleration is not (yet) a useful concept and therefore has not been given a name, you are correct that whatever it is called (lets call it Eric for the sake of sanity) it will be the derivative of Velocity with respect to time.

Eric = [symbol]d[/symbol]v/[symbol]d[/symbol]t

:slight_smile:

The third derivative of the position function is called jerk. That’s pretty standard.

Beyond that, it’s not standard, but it’s not uncommon to see references to the following (which were created half-jokingly):

4th derivative: snap
5th derivative: crackle
6th derivative: pop.

http://math.ucr.edu/home/baez/physics/General/jerk.html

change in something is often called delta something. the delta is a triangle with point upward; you know, the greek letter. i don’t know how to make one appear. perhaps grimpixie can enlighten me, since it knows how to make a lowercase delta.

so change in acceleration can be called delta a or ^a (just pretend that’s a proper delta).

i don’t think it’s strictly proper to say delta acceleration or delta velocity, but all my physics profs have said delta v or similar (rate of change of acceleration is less common).

jim, i don’t think calculus will give you the term you are looking for, but it could certainly give you derivitive (and integral) equations with regard to t…as grimpixie said, i see.

In my high school physics lecture I was told that the common word for the change of acceleration is Impulse. Don’t know if it’s scientifucally correct but we used it in class…

Impulse is (according to my notes) defined at force multiplied by time.

Also, delta acceleration would be change in acceleration over time, while the derivative in question would be the instantaneous rate of change.

– CH

I agree with jerk

It is definitely jerk. I was taught that by at least 3 physics teachers. :slight_smile:

IIRC correctly, impulse is a delta of momentum.

It is a useful concept when dealing with rockets and jet engines. The Argus engine on the V-1, for example, had increased thrust with increased velocity, so the accel of accel is useful. It’s also good for diagnostics in amateur rocketry; you can get an idea of how the grain is burning by how the acceleration changes.

I’ve never heard it before, but jerk sounds good to me.

A calculus teacher told us it was called warp, but maybe he was just a Star Trek geek…

jerk:
a unit of change in acceleration sometimes used by engineers. This is not as silly as it sounds, because when we’re in a vehicle and feel a jerk, we are in fact experiencing a change in the acceleration of the vehicle. One jerk is equal to a change in acceleration of one foot per second per second in one second, that is, 1 ft/sec3. One jerk equals 0.3048 m/s3 or about 0.03108g /sec.

http://www.unc.edu/~rowlett/units/dictJ.html

Strange to hear it described as a unit. I have heard it described not as a unit but as a general concept, as described here:

What is the term used for the third derivative of position?

I guess there isn’t wrong with it being a unit also, esp since as an english system unit it is pretty worthless anyway. :slight_smile:

This could make a good Abbott and Costello routine. “What is change in acceleration called?” “Jerk.” “No need to call me names, I was just asking!” :smiley:

That said, I don’t remember any ballistics textbooks/problems I’ve seen giving this a name. It’s usually just referred to as da/dt. (And it really does come up a lot-even with a constant force behind it, the acceleration of a rocket increases over time as the fuel is spend and the total mass decreases …)

If you can’t see this post, I’ve given up after three tries.

We were told that the jerk is what the human body is most sensitive to – perhaps the only motion that can be felt internally. Steady acceleration (cf. Einstein’s elevator) is difficult to feel without aural or visual clues, but when you’re getting jerked around, you know it. It also has application in looking at sports injuries.

If I’d been the first to respond to this question, I would have had a strong temptation to just leave the response, “jerk”.

So the cardinal rule of the SDMB is “Don’t be a d[sup]3[/sup]s/dt[sup]3[/sup].”

Interesting stuff. I was a physics major in college, and I have often wondered if there was a word for that. I even asked a prof once, and he didn’t know. The same professor said that “pickup” is the scalar equivalent of vector acceleration. Was he full of it?2

Hey, bibliophage you may want to get one of these (from Cabbage’s link):

Might make your job easier. :slight_smile:

Oh that just screams sig line to me.
:cool:, bibliophage

bibliophage, that is so beautifully geeky! :smiley:

For the sake of correctness, the value we’re talking about is the second derivative of velocity with respect to time, not the first. It is the first derivative of acceleration, however.

Jerk! That’s actually a new one for me, but it sure is neat. As for the impulse issue, you guys are both right- it’s force times change in time and change in momentum.

Is there a way to insert a delta on this group? Hmm? Well, can’t find one, so let d(x) read ‘delta x.’ Quick proof:

F = ma
F = md(v)/d(t)
Fd(t) = md(v)

Change in momentum equals force times the time over which the force acts. Cheers! (In the past two weeks I’ve taken a physics final and a calc final. Now begone!)
F = d(p)/d(t)