11

Is it possible if its motion were discontinuous?

I'm trying to understand if there is a function that has this property, but I chose to say it in terms of motion because it's easier to explain.

Mostafizur Rahman
  • 434
  • 2
    It seems that people have vastly misinterpreted this question. To put it mathematically : Let $f$ be a function such that $f(0)=0$ , and such that $f(x) - f(x-3) = 30$ for any $x \geq 3$. Is it then true that $f(x) - f(x-2) = 20$ for any $x \geq 2$? The accepted answer does the job, I think. – Sarvesh Ravichandran Iyer Jul 15 '22 at 06:34
  • The motion of a particle with mass could not be discontinuous. The particle would exist at the same time at two different places at every discontinuity. Speed would be infinity. – Uwe Jul 15 '22 at 14:32
  • 3
    What I particularly like about this question is that the answer is not only 'yes, there is such a function', but in fact almost any function will have this property. Choose any non-constant function on the interval $[0,3)$, and make it repeat every three seconds. To make things continuous make it a continuous function with the same start and end speeds. Make sure its integral is non-zero so that you actually go somewhere, and normalize it so that you travel the right distance. Make sure it isn't a function on $[0,1)$ that is repeated and it should break the 2 second rule. – David A. Craven Jul 15 '22 at 19:19
  • 1
    Are you asking if there is at least one two second interval with that property, or if every two second interval will have that property? The latter question is basically trivial, but that seems to be how every answer interpreted it. – BlueRaja - Danny Pflughoeft Jul 16 '22 at 00:33
  • @BlueRaja-DannyPflughoeft: It seems that both questions are fairly trivial, but maybe not so to the OP. OP eventually seemed to indicate that they meant the latter (in their comment to the accepted answer). – Brian Tung Jul 20 '22 at 18:25

5 Answers5

20

It isn't true, not even for continuous motion.

For example, let $V(t)=10+\cos\left(\dfrac{2\pi}{3}t\right)$ in meters/sec.

Then

\begin{eqnarray} D&=&\int_0^3V(t)\,dt\\ &=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_0^3\\ &=& 30 \end{eqnarray}

but

\begin{eqnarray} D&=&\int_0^2V(t)\,dt\\ &=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_0^2\\ &=& 20+\frac{3}{2\pi}\sin\left(\dfrac{4\pi}{3}\right)\\ &=&20-\frac{3\sqrt{3}}{4\pi} \end{eqnarray}

For any three second time interval the distance traveled will be 30 meters because of the period of the sinusoidal portion of the velocity.

ADDENDUM: In response to a comment from @TheRubberDuck

\begin{eqnarray} D&=&\int_x^{x+3}V(t)\,dt\\ &=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_x^{x+3}\\ &=& 10(x+3)+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}(x+3)\right)- \left(10x+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}x\right)\right)\\ &=&30+\frac{3}{2\pi}\left[ \sin\left(\frac{2\pi}{3}x+2\pi \right)-\sin\left(\dfrac{2\pi}{3}x\right) \right]\\ &=&30 \end{eqnarray}

However, integrating between $x$ and $x+2$ gives a factor

\begin{eqnarray} \sin\left(\frac{2\pi}{3}x+\frac{4\pi}{3} \right)-\sin\left(\dfrac{2\pi}{3}x\right) \end{eqnarray}

which is not identically $0$.

John Wayland Bales
  • 21,814
  • 5
    "For any three second time interval the distance traveled will be 30 meters because of the period of the sinusoidal portion of the velocity.": I think it would be good to show that by integrating from $x$ to $x+3$ instead of $0$ to $3$. – TheRubberDuck Jul 15 '22 at 15:30
  • @TheRubberDuck I will add a bit to my answer to make that clearer. – John Wayland Bales Jul 15 '22 at 15:36
  • 2
    "which is not identically 0" - I think the relevant question is not whether it's identically zero, but whether there are any 3-second intervals during which that extra factor is always > 0. To which the answer is no, because $\sin\left(\frac{2\pi}{3}x+\frac{4\pi}{3} \right)-\sin\left(\dfrac{2\pi}{3}x\right) = A\cos\left(\frac{2}{3}(\pi x+1)\right)$This has a period of 3, aka a zero every 1.5. Meaning for any three second period, this function will always have a two-second period where it travels at least 20. – BlueRaja - Danny Pflughoeft Jul 16 '22 at 00:47
  • @BlueRaja-DannyPflughoeft Clearly there will be a specific two second interval during which distance traveled is 20 meters, but I believe OP was asking whether that would be the case for any 20 second interval whatsoever. – John Wayland Bales Jul 16 '22 at 00:47
  • @JohnWaylandBales The OP asked "does it ... in any 2 seconds?", not "does it ... in every 2 seconds?". So if it can be proved that there is any 2-second interval where it travels 20 metres, the answer is "yes". – Rosie F Jul 16 '22 at 05:10
  • 2
    @RosieF The word 'any' is ambiguous. But I am assuming that OP is using it in the same sense in the last half of the statement as in the first half. If that is not the case, then perhaps OP can clarify. – John Wayland Bales Jul 16 '22 at 05:27
  • 1
    I should have said "every". Pardon my bad English. – Mostafizur Rahman Jul 19 '22 at 17:15
12

No. Imagine a body that is stationary for two seconds, then moves $30$ meters in the next second, and then repeats this indefinitely. Then in any three-second interval, it moves $30$ meters, but there are clearly two-second intervals in which it doesn't move $20$ meters (in which, in fact, it doesn't move at all).

$$ x(t) = \begin{cases} 30k & 3k \leq t \leq 3k+2 \\ 30k+30[t-(3k+2)] & 3k+2 \leq t \leq 3k+3 \end{cases} \qquad k \in \mathbb{Z} $$

I've given a function that isn't differentiable, but it could be made differentiable quite easily.

Brian Tung
  • 34,160
  • 1
    During the interval $[1, 4]$, the particle travels 30 meters. It's perfectly possible that it travels 13 meters during the interval $[1, 3]$ and 17 meters during the interval $[3, 4]$, making it clear that it doesn't necessarily travel 20 meters in any 2 seconds. But, when I say, the particle travels 30 meters in any 3-second intervals, it also includes the intervals $[1.23, 4.23], [1.61803, 4.61803], ...$ So, it seems to me, that your logic accounts for only a certain type of infinitely many 3-second intervals. – Mostafizur Rahman Jul 15 '22 at 06:20
  • 1
    @MostafizurRahman: I don't believe that's right. For instance, for the interval $[1.23, 4.23]$, we have $x(1.23) = 0, x(4.23) = 30$ (I miswrote the symbolic form, but I'll fix that now). It travels $30$ meters in those $3$ seconds, and indeed, it is always the case that $x(t+3) = x(t) + 30$ for all $t$, and yet it is not the case that $x(t+2) = x(t) + 20$ for all $t$. – Brian Tung Jul 15 '22 at 07:34
  • A particle with mass could not move according to a not differentiable function, speed and acceleration would be undefined. – Uwe Jul 15 '22 at 14:04
  • 1
    @Uwe: True, but it's not difficult to make this differentiable (replace the straight rising segment with a sigmoid-type on compact support), and the OP has already stated that this is just a proxy for a more abstract question regarding the function. – Brian Tung Jul 15 '22 at 15:21
  • 1
    Concentrating the movement in short bursts seems like a way to generate many functions that satisfy the criterium in the question. I'm wondering if it's even possible to construct a function that way which has an even stronger property: there's no 2 second interval in which the particle travels 20 meters. My first intuition is no, though, because the speed has to be on average 10 m/s and continuous. – MSalters Jul 15 '22 at 21:01
  • 2
    @MSalters: I came to ask something similar. Couldn't the mean value theorem be used in order to prove that if a particle travels 30m in 3s, there must be at least one 2s interval during which the particle travels exactly 20m? – Eric Duminil Jul 15 '22 at 21:43
  • @EricDuminil: If the pattern repeats every $3$ seconds, then yes. If you only look at a single $3$-second segment, then possibly not. – Brian Tung Jul 20 '22 at 18:19
  • @MSalters: As David Craven indicates in his comment to the OP, most functions will work. – Brian Tung Jul 20 '22 at 18:24
  • @BrianTung: Thanks. Do you happen to have such example? – Eric Duminil Jul 20 '22 at 18:33
  • 1
    @EricDuminil: A single three-second segment in which the particle advances $15$ m in $0.5$ s, and then doesn't move for $2$ s, and then advances another $15$ m in the last $0.5$ s, never moves exactly $20$ m in any two-second interval. If that pattern were to repeat, however, then it would indeed move exactly $20$ m in some two-second intervals—for example, between $t = 7/6$ and $t = 19/6$: $x(7/6) = 15$ and $x(19/6) = 35$. – Brian Tung Jul 21 '22 at 00:29
2

This (30m in 3sec and 20m in 2sec) will only be applicable for motion with constant speed (ie. Linear function)

But if the motion is accelerated (ie. Degree of function is more than 1) Or there are jerks in motion (ie. Function is discontinuous at some point(s)) then covering the given distance is possible in other ways.

For example when a person is traveling by car and reaches the destination in some time t then it is possible that he may have stopped the car somewhere or slowed due to traffic and may have had more speed at some other points. While another person travelled the same distance at constant speed and also reached at the same time t.

Salvik
  • 37
  • 2
0

What does it mean for the motion to be discontinuous?

If you are talking about the distance-time graph, it cannot be discontinuous.

${\text{The main thing}}$:

If a body travels $30$ meters in any $3$, it is not necessary that it travels $20$ meters in any $2$ seconds. For example, a body travels $5$ meters in the first second, $10$ in the second and $15$ in the third. Then $5$ again, then $10$ again, then $15$ again and so on.

See this:

$$5,10,15,5,10,15,\cdots$$

F. A. Mala
  • 1
  • 1
    I said it travels 30 meters in any 3 seconds. That is, in all the infinitely many 3-seconds intervals, it travels 30 meters. So does this alone make its motion uniform? – Mostafizur Rahman Jul 15 '22 at 05:50
  • @mostafizur

    Kindly see it again.

     – F. A. Mala Jul 15 '22 at 05:57
  • 1
    During the interval [1,4], the particle travels 30 meters. It's perfectly possible that it travels 13 meters during the interval [1,3] and 17 meters during the interval [3,4], making it clear that it doesn't necessarily travel 20 meters in any 2 seconds. But, when I say, the particle travels 30 meters in any 3-second intervals, it also includes the intervals [1.23,4.23],[1.61803,4.61803],... So, it seems to me, that your logic accounts for only a certain type of infinitely many 3-second intervals. – Mostafizur Rahman Jul 15 '22 at 06:24
  • 1
    I think you are right. – F. A. Mala Jul 15 '22 at 06:32
-1

Suppose you have a vehicle traveling 10 m/s, and suppose the wheels have a diameter such that they make a complete revolution every three seconds. Then after three seconds, every point on the wheels will be at the same position relative to the vehicle, but will have moved 30m. On the other hand, after two seconds, points on the wheels will be in a different position, so they will have moved 20m plus or minus whatever distance they traveled relative to the vehicle.

Acccumulation
  • 12,210