If there exists something like that, then in $distance/time/time/time$, how is it expressed?
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3$\begingroup$ Related: physics.stackexchange.com/q/136880/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Apr 11, 2015 at 19:29
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2$\begingroup$ In fact, it is not only possible, it is necessarily the case. That is to say, uniform acceleration is not physically realizable. $\endgroup$– Alfred CentauriCommented Apr 11, 2015 at 21:58
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4 Answers
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http://wordpress.mrreid.org/2013/12/11/jerk-jounce-snap-crackle-and-pop/
Speaking derivatives to time:
- First position $x$,
- then velocity $v=x'=\frac{dx}{dt}$,
- then acceleration $a=x''=\frac{d^2x}{dt^2}$,
- then jerk $x'''=\frac{d^3x}{dt^3}$,
- then jounce/snap $x''''=\frac{d^4x}{dt^4}$,
- then crackle $x'''''=\frac{d^5x}{dt^5}$,
- then pop $x''''''=\frac{d^6x}{dt^6}$,
- ...
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4$\begingroup$ Formally, the fourth-derivitive of position is the jounce, however, someone published a paper using snap (see also the footnote on page 4). On a less serious note, when multiplied by mass, you get, respectively: kilogram-meter ($m\cdot x$), momentum ($m\cdot v$), force ($m\cdot a$), yank ($m\cdot jerk$), tug ($m\cdot snap$), snatch ($m\cdot crackle$), shake ($m\cdot pop$). $\endgroup$ Commented Apr 11, 2015 at 22:56
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2$\begingroup$ In valvetrain design engineering areas of linear snap (constant pop) are used as design limits for valve lift profiles. $\endgroup$ Commented Apr 12, 2015 at 0:02
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1$\begingroup$ d4x/dt4, d5x,/dt5, d6x/dt6, Rice Crispies! $\endgroup$ Commented Apr 12, 2015 at 7:17
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Yes. Rate of change of acceleration is called jerk. Yes its dimensional formula is $[M^0, L^1, T^{-3}]$. Similarly one could also define higher time derivatives of acceleration if required for a particular problem.
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3$\begingroup$ There is a name for the further derivatives of acceleration too, i believe jounce is the second derivative of a, though after that its not a consensus among physicists $\endgroup$ Commented Apr 11, 2015 at 22:46
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Yes. usually we name them $a'$ . and there can be even a speed of my $a'$ that I can call that $a''$ and it goes on like that.
it is only used it real life calculation that the calculation should be very precise like rocket science.
and the equation of displacement (with a constant $a'$) will be :
$$x =\frac16 a't^3 + \frac12 a_0t^2 + v_0t+x_0$$
(EDIT: also should mention sometimes they put a little dot on the a too instead of apostrophe )
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2$\begingroup$ You might want to add that the equation holds only for constant rate of change of acceleration. $\endgroup$ Commented Apr 11, 2015 at 19:31
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$\begingroup$ Regarding the edit - yep, $\dot{b}$ indicates the change in a quantity, $b$, with respect to time: $\frac{db}{dt}$. $\endgroup$ Commented Apr 11, 2015 at 20:05
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$\begingroup$ @HDE226868 thanks. I remember one my professors mentioned that you use apostrophe (and he always used this method) to mention the second acceleration. $\endgroup$– MobinCommented Apr 11, 2015 at 20:08
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An actual example in which there is a non-zero change in acceleration, that is, jerk, occurs is a spring. A spring's motion is described by a sinusoidal function. The derivative of a sinusoidal function is just another sinusoidal function. As a result, you can differentiate such a function infinitely many times, and will never have a derivative that's 0/a constant. So not only is there there a non-zero jerk in the motion of a spring, every single derivative of position is non-zero.
answered Apr 12, 2015 at 4:52