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Today, I encountered the problem where I was asked to find the velocity and acceleration from displacement-time graph but the displacement-time graph was discontinuous. So I am unable to find the velocity where it was discontinuous.

I know from my Calculus class that we cannot find the derivative when function is discontinuous at some point.

So I think it's not possible to find the velocity and acceleration at that point. So what is the physical significance of that?

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    $\begingroup$ We need better information. Do you mean you have an equation which produces a discontinuity, or just that you have a set of data points with a big jump at some sample point? $\endgroup$
    – Carl Witthoft
    Commented May 5, 2020 at 14:03
  • $\begingroup$ Yes! Actually it was graph... Not a equation... So it's quite clear from the graph that it was discontinuous. $\endgroup$
    – Roger Michealson
    Commented May 5, 2020 at 15:34
  • $\begingroup$ So you were asked to find the acceleration of a teleportation? $\endgroup$
    – Mehdi
    Commented May 6, 2020 at 10:08

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In the real world, a displacement time graph can never be discontinuous. The only not-so-physical meaning that it has is that the body teleported from one place to another such that its displacement changed instantaneously/discontinuously. And as you can see, it's impossible to define the velocity of this teleportation.

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    $\begingroup$ Discontinuously does not mean that body has moved instantly, it can simply mean that you don't have enough data points to describe body movement fully. I.e. consider you see a car in a night moving from point A to B. Then at point B car switches-off it's lights, and you loose it form your sight. Then car turns-on lights at point C and you follow it to point D. In this case you will have observation discontinuity B-C, but it does not mean that car traveled instantly this sector, just that you didn't saw how it moved. $\endgroup$
    – Agnius Vasiliauskas
    Commented May 5, 2020 at 15:06
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    $\begingroup$ @AgniusVasiliauskas: In that sort of situation, the position function $x(t)$ is undefined for some range of $t$. This is distinct from what I usually think of (and I suspect many people think of) as a "discontinuous function", for which a position $x(t)$ is defined for all $t$ but for which a derivative of $x$ does not exist at some values of $t$. $\endgroup$
    – Michael Seifert
    Commented May 5, 2020 at 15:20
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    $\begingroup$ @AgniusVasiliauskas I exactly second Michael Seifert's comment. I assume that the OP is talking about a case when the function is defined everywhere but has at least one point where the limit is not defined. $\endgroup$
    – user258881
    Commented May 5, 2020 at 15:32
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    $\begingroup$ Yes! The graph is discontinuous at only one point. $\endgroup$
    – Roger Michealson
    Commented May 5, 2020 at 15:36
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    $\begingroup$ @AgniusVasiliauskas a discrete sample is technically speaking always everywhere discontinuous, so that interpretation doesn't make much sense. Yes, it may in practice happen that there's a much bigger gap in some region, but then it's already evident that it's not meaningful to talk about derivatives there – nor about function values themselves, for that matter. $\endgroup$
    – leftaroundabout
    Commented May 5, 2020 at 16:38
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@Fakemod has answered your question perfectly. But I'd like to add an extra point. You should also know that you cannot apply calculus if the graph is non differentiable at a point, not necessarily discontinuous.

In such cases, the simple explanation is that the equations describing the motion have changed, hence, it won't make sense to use a direct integral.

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    $\begingroup$ You can apply some parts of calculus, but yes certainly you can't provide a value for the derivative at a point which is non-differentiable. $\endgroup$
    – Carl Witthoft
    Commented May 5, 2020 at 14:04
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You can extrapolate average speed from a continuous regions separated by discontinuities.
Take a look at such function jump discontinuity case :

enter image description here

In this case body average speed would be :

\begin{align} v &= \frac {dx(t)}{dt} \\&= \frac 12 \left( \sin(t)^\prime + (2t)^\prime \right) \\&=\frac{\cos(t)+2}{2} \end{align}

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  • $\begingroup$ That is mathematically a discontinuity, but in physics it's just an "unknown/ no data" region. I think the OP is claiming he has two $x$ values for a single $t$ value in his graph. $\endgroup$
    – Carl Witthoft
    Commented May 5, 2020 at 18:40
  • $\begingroup$ I've fixed scheme to be real function jump discontinuity case. There are several cases of discontinuities - removable discontinuity, jump discontinuity and essential discontinuity. I've quite sure that my idea of function averaging works even for essential discontinuity type. Question is only, how much useful such averages are. For me, it has usefulness, because by averaging, you make a global "net" continuous function. $\endgroup$
    – Agnius Vasiliauskas
    Commented May 5, 2020 at 19:47
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    $\begingroup$ It makes no sense to calculate the average speed the way you've done. The instantaneous velocity at time $t$ is either $\sin t$ or $2$, not some arbitary average of both. And if you want to calculate the average velocity over some time interval, you can do that, but you need to specify the starting and ending time with more detail than you've done. $\endgroup$
    – Michael Seifert
    Commented May 6, 2020 at 0:24
  • $\begingroup$ But as long as we don't know which of both values speed will take,- it makes sense to take the average as good enough approximate to both functions at discontinuity, yes? No other way around to solve this issue i think. And yes, we may add additional constraints, re-scale functions, translate them and do whatever you see fit to them. No one solution for all. $\endgroup$
    – Agnius Vasiliauskas
    Commented May 6, 2020 at 0:38
  • $\begingroup$ Btw, in case of essential discontinuity - where there are many points of $t$ with multiple values of $x$ - specifying time range bounds exactly will not help you at all. You will still need to do some sort of function composition and/or averaging to solve discontinuities issue. $\endgroup$
    – Agnius Vasiliauskas
    Commented May 6, 2020 at 5:50
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Numerically assume then explicitly state a '$dT$' that equates to the smallest duration of time that you can ascertain from the graph/data (perhaps does your graph have a grid? If so, use the duration of a single step in time). Then calculate the difference between the discontinuous points and divide by $dT$.

Just be sure to explain it, even if your guess for $dT$ is silly you should get points for making the best of an undefined situation 👍

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