# When we take time derivative of a function of time, then is the result another function of time, again?

(I'll try to explain my question by one known example), for example where the velocity is a function of time v(t) then its time derivative (which is acceleration: $a=\frac {dv}{dt}$) is another function of time a(t)? (because according to defination of integral it must be another function of time: $v=\int a(t)dt$) and then what will be time derivative of acceleration?

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What if $v=kt$, so $a=k$? –  jinawee Jun 17 '13 at 11:38
Note that there is such a thing as a constant function: a function whose value happens to be a constant independent of the value of the independent variable. Such a thing is still a perfectly good function. :) –  Michael Brown Jun 17 '13 at 11:58
Time derivative of acceleration is called jerk, and then snap crackle and pop for further derivatives. Look it up. –  ja72 Jul 9 '13 at 13:53
Relevant wikipedia link: en.wikipedia.org/wiki/Jerk_(physics) –  ja72 Jul 9 '13 at 14:01

You've obviously seen the equation $v = \int a(t)dt$ somewhere and been mislead by it. It's entirely possible for the acceleration to be a constant in which case we get $v = \int k \space dt$ for some constant $k$, and therefore $dv/dt$ is a constant, $k$, and not a function of time. We write the acceleration as $a(t)$ because this is the most general form and covers all eventualities.

The time derivative of acceleration is known as jerk. In the example above, where $a$ is constant, the jerk will be zero.

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In general the acceleration would be a function of time, i.e. $d^2x/dt^2 = a(t)$.

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You've probably seen the equation for position when the acceleration is the constant $a$:

$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$

Differentiating once yields:

$\dot x(t) = v(t) = v_0 + at$

Differentiating again yields:

$\ddot x(t) = a(t) = a$

Now, it may seem funny to write $a(t) =$ constant but it's perfectly acceptable because, in general, acceleration is a function of time. In this case, it just happens to be a function of $t^0$.

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The derivative of acceleration with respect to time is called jolt, or jerk. Sometimes, this jolt (jerk) may be non-differentiable so that the jounce (derivative of jolt/jerk) is not continuous. In such cases, the fourth, or more generally, the $n^{\operatorname{th}}$ derivative of the position may not be continuous. Some times, it is differentiable an arbitraryily large number of times, and then, for the position $x(t)$, a polynomial function of time $t$, differentiating it $n$ times results in a constant value, where $n$ is the degree of the polynomial function of time $t$ which the position $x$ is.

So, it is easier to write $\frac{d^nx}{dt^n}$, and solve the differential equation you get. Hope it helps!

(If you believe I have misinterpreted your question, kindly mention so in the comments)

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Your English is very unclear. dimension10's edit was fine (I couldn't spot any errors - he never said that continuity implies differentiability), but this version is very difficult for a native English speaker to understand. –  Michael Brown Jul 9 '13 at 11:10
Yes, I understand the difference between continuity and differentiability. I think you are misunderstanding what dimension10 said. He was saying that the derivative of a non-differentiable function is discontinuous, a statement which can be made rigorous and is plainly true for any function which is only non-differentiable pointwise. That is, for any function the OP will ever come across (unless they start doing path integrals). It would be best to write in clear grammatical English from the start to avoid these confusions. I wish it weren't so, but alas, English is the lingua franca. –  Michael Brown Jul 9 '13 at 14:16
@SaurabhRaje: If you think that the only thing necessary to edit was grammar, then why don't you rollback to my revision, since there the grammar was also correct and the content was the same? P.S. If you noticed, you got an upvote after I edited, and a downvote after you reversed my edit, which means that I'm not the only one who can't understand what you have written. I won't reverse your rollback, because I don't want to cause an edit war, but if you want, you can rollback it to my edit (v2). Thank you if you will, and I'm downvoting for now. –  Dimensio1n0 Jul 15 '13 at 9:43
As a native speaker, I find your answer understandable and there's no need to edit it. –  Larry Harson Jul 15 '13 at 13:41
@LarryHarson: Yes, I think it's fair. I have not personally attacked anyone. Just said, "Nobody can understand. " And do you really think it is reasonable to upvote this? SaurabhRaje: Physics.SE is not a battlefield. If you can't take something like "It was not understandable. ", then it is no wonder that you are uncomfortable with people pointing out flaws with your physics reasoning (on other questions. ) . –  Dimensio1n0 Jul 15 '13 at 18:39