# Non-zero higher time derivatives of position?

My mom told me to use speed control, which would allow the car to remain at constant speed. I told her that its impossible for a car to maintain constant speed, as slight changes in friction on the road cause differences in the acceleration of the car, and hence the velocity.

But that got me thinking: Even the differences in acceleration caused by differences in the friction due to the road need not be constant. Thus, the car's acceleration is also always changing, implying that it has non-zero "jerk," defined as the change in acceleration over time.

Is this reasoning correct? If so, is there anything keeping the "jerk" constant? How about snap, crackle, and pop (the 4th, 5th and 6th time derivatives of position)? Does it even have to stop there?

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There is no reason your displacement couldn't be an exponential function, or non-positive integer power-law (for some period of time) with an arbitrary number of non-zero derivatives. –  zhermes Jan 17 '13 at 2:35
This is exactly what I wanted to hear. You clearly understood my question. +1 –  chubbycantorset Jan 17 '13 at 7:01

Forget the friction part, unless you're always peeling rubber or screeching the brakes.

The acceleration is the combined sum of forces - air drag, engine thrust at the wheels, slope of the road, etc., divided by mass (weight). You're right. Those things change all the time, so the acceleration and speed are always changing. The cruise control is a feedback system that tries to adjust the throttle to stay at one speed, but only approximately.

Cars are designed with a property called driveability, which basically means they're not too jerky in acceleration, so they do try to limit the 3rd derivative. The 2nd derivative, acceleration, is limited by engine power and the amount of available tire friction. Forget the higher derivatives. You can take as many derivatives as you want. It's a continuous system and infinitely differentiable.

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My question isn't asking about how the car measures the velocity. I'm only interested in what kind of a function acceleration can be in this case, i.e. if there exists an $n$ for which $f_a \in C^n$, where $f_a$ is the acceleration function. –  chubbycantorset Jan 17 '13 at 0:36