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A car starts distance 1 from a wall then drives away at constant acceleration $c$. There is a length of elastic tied between the wall and the car. Remarkably, this doesn't affect the motion of the car (in reality the elastic would snap or its tension slow the car).

Meanwhile, an ant crawls from the wall on to the elastic then towards the car. On solid ground, the speed of the ant is $a$.

Will the ant ever reach the car? Assume the elastic can stretch infinitely.

(Note: Assume that the laws of Newtonian mechanics apply, and that there are no effects with relativity.)

This question is related to a previous puzzle, posted by Colonel Panic.

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    $\begingroup$ @SinanSamet, this is the second part of that question. $\endgroup$
    – Kenshin
    Nov 10, 2014 at 14:26
  • $\begingroup$ But how? It's the exact same question by another user so what stops me from making a part 3 with the exact same question? I'm new here so I might be misunderstanding this, and if so sorry for that it just seems weird to me. $\endgroup$
    – Sinan Samet
    Nov 10, 2014 at 14:31
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    $\begingroup$ @SinanSamet note the bold acceleration in the question. That's the difference that makes it a (very) different question. $\endgroup$
    – frodoskywalker
    Nov 10, 2014 at 14:38
  • $\begingroup$ Oh wow sorry for my stupidity and blindness I didn't notice that. $\endgroup$
    – Sinan Samet
    Nov 10, 2014 at 14:40
  • $\begingroup$ A car starts distance 1 from a wall... Does the ant's step cover more or less distance than (1)? $\endgroup$
    – user2338816
    Nov 11, 2014 at 13:06

6 Answers 6

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The ant never reaches the car, if the car is fast enough.

At time $t$, the velocity of the car is $ct$, and the length of the rope is $1+\frac12ct^2$. Track the fractional progress of the ant, who starts 0% of the way along the rope and wishes to reach the car positioned at 100%.

At time $t$, the ant's fractional speed is inversely proportional to the length of the rope. It is $\frac{2}{2+ct^2}a$.

So the question is really: is $\int_0^\infty\frac{2}{2+ct^2}a\text{ }dt$ at least 1? We could evaluate the integral using the arctan function, but there is an easier way to proceed. Recall that $\sum_0^\infty\frac{2}{2+ct^2}$ converges. Therefore, by the integral test for infinite series, $\int_0^\infty\frac{2}{2+ct^2}a\text{ }dt$ is finite. Therefore, if $a$ is small enough, then the integral is less than 1, and the ant never reaches the car.

EDIT: By request, we can figure out the exact conditions for the ant reaching the car. By integrating using the arctan function (or WolframAlpha), we find $\int_0^\infty\frac{2}{2+ct^2}a\text{ }dt=\frac{\pi a}{\sqrt{2c\text{ }}}$. The ant reaches the car iff this quantity is greater than 1.

Therefore, the ant reaches the car iff $a>\frac{\sqrt{2c\text{ }}}{\pi}$.

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  • $\begingroup$ That's what I would have said. Next question is: what is the relation between c and a so that the ant reaches the car? $\endgroup$
    – Florian F
    Nov 10, 2014 at 17:39
  • $\begingroup$ @FlorianF: I've edited the answer to include the relation between c and a. But I've never seen an ant who can crawl at $\frac{\sqrt{2c\text{ }}}{\pi}$ meters per second. :) $\endgroup$
    – Lopsy
    Nov 10, 2014 at 19:17
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    $\begingroup$ How interesting that pi comes out of a puzzle that has nothing to do with circles. $\endgroup$
    – Kenshin
    Nov 11, 2014 at 7:04
  • $\begingroup$ $a$ and $\frac{\sqrt{2c}}\pi$ have different units. $\endgroup$
    – user253751
    Nov 12, 2014 at 4:37
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    $\begingroup$ @immibis Ooh, that worried me until I worked through the integral again. The 1 in the numerator of the integrand is actually "1 meter". So the $\sqrt{2c}$ in the end turns out to be meters per second, because the 2 is actually 2 meters. So the formula works as long as you use m/s. $\endgroup$
    – Lopsy
    Nov 13, 2014 at 14:53
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Yes.

There are 3 possibilities for constant acceleration; Constant positive acceleration, zero acceleration, constant negative acceleration. If the car were already moving and we are using a negative acceleration then the car would eventually stop and the acceleration would no longer be constant so this isn't the case.

If the acceleration were positive the car would eventually reach the speed of light which should be impossible even in this universe where there is no force associated with an infinitely elastic rubber band and cars can run infinitely without recharging or refueling.

Therefor the car's acceleration is 0 which is constant and since it has always been zero the car has not moved and the ant will make it his speed of a or it was already moving and the problem becomes the same as the one this question was derived from

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  • $\begingroup$ interesting answer, but I have edited to assume that in this universe there is no maximum speed limit. $\endgroup$
    – Kenshin
    Nov 10, 2014 at 14:27
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    $\begingroup$ This is not correct. You can accelerate with a constant positive acceleration in our universe. After a long enough time, you will approach arbitrarily close to the speed of light from the perspective of a stationary observer, but as far as you are concerned, you are nowhere near the speed of light. $\endgroup$
    – Nick Matteo
    Nov 10, 2014 at 20:30
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    $\begingroup$ If your acceleration is negative, you do not stop accelerating when velocity reaches 0. The acceleration remains constant and you increase velocity in the opposite direction. $\endgroup$
    – eclipz905
    Nov 10, 2014 at 23:13
  • $\begingroup$ If the acceleration is negative the ant is eventually going to be squashed between the back of the car and the wall. $\endgroup$
    – Penguino
    Nov 12, 2014 at 3:55
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This should probably be on the math stackexchange. It's a differential equations problem.

The interesting part about the elastic band is that it provides the ant with a fraction of the car's velocity. The fraction varies depending on how far along the elastic the ant has gotten. For instance, at t = 0 when the ant is at the very beginning, the elastic is anchored and doesn't move. But at t = ? when the ant is halfway to the car, the elastic point is moving at half the speed of the car. And at t = ? when the ant catches up, the elastic is moving at the full speed of the car.

The distance of the car from the wall at time t is $d_c = \frac{1}{2} c t^2 + 1$

Let the distance of the ant from the wall at time t be $d_a$, which is a function of t

The velocity of the ant at time t is $d_a'$ (first derivative of $d_a$)

We know that $d_a' = a + \frac{d_a} {d_c} d_c'$ (the constant component a, plus a fraction of the car's velocity based on how far along the elastic the ant is at the time)

Substituting: $d_a' = a + \frac{d_a} {(\frac{1}{2} c t^2 + 1)} (c t)$

That's a first order differential equation. I plugged it into Wolfram Alpha's differential equation solver. It provided a complicated solution:

$y = \frac{a(c t^2 + 2)\arctan(x\sqrt{\frac{c}{2}})}{\sqrt{2c}} + k_1(c t^2+2)$

$k_1$ is a constant that we have to provide based on the initial conditions. Our initial condition is y(0) = 0. Substituting that in, we see $k_1$ must be 0.

So we're left with $y = \frac{a(c t^2 + 2)\arctan(t\sqrt{\frac{c}{2}})}{\sqrt{2c}}$

The ant catches up to the car when $d_a = d_c$, or

$\frac{a(c t^2 + 2)\arctan(t\sqrt{\frac{c}{2}})}{\sqrt{2c}} = \frac{1}{2} c t^2 + 1$

That's too hairy to keep working with. But there are definitely solutions. Pick an x (time), a c (car's acceleration), substitute, and solve for a, the required minimum speed of the ant. For instance, t = 10, c = 1, results in a = ~0.494.

Hopefully someone can check my math, because I'm not sure I did all the math markup correctly!

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NO, if the car accelerates fast enough!

If the ant travels at a speed of a m/s If the car starts at a speed, which will strech the rubberband in front of the ant by a length of a/s, the distance of the ant to its goal will not increase. If the and comes closer to its goal, the car has to speed up.

So if the ant has reached halfway on the rubber-band the car would have to drive twice as fast, so it will still strech the band in front of the ant by a m/s, so it has to strech the whole band by 2 a m/s So if the acceleration is large enough (or more) to get the car to double its speed in the same time the ant reaches 50%, the ant will never catch up, since the rubberband in front of the ant is always expanding faster than the ant can travel, so the distance to the car always increases (in the edge case actually stays the same)

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The ant reaches the car if the distance of the ant from the wall is less than distance of car from wall.

Or

dant <= dcar

Using basic calculus rules of integration, and knowing the relationship between acceleration, speed, and distance we have the two equations:

dant = a*t

dcar = 1/2*ct^2 + st + 1

Where t is the amount of time passed, and s is the initial speed of the car.

Thus, the ant will reach the car in situations where

a*t <= 1/2*ct^2 +st + 1

Where a and c (and s) are any real numbers. As the poster above mentions, acceleration can be positive negative or zero. If the acceleration is negative or zero, then situation becomes trivial. The ant's speed just has to be greater than zero. If the acceleration is positive, then the relationship between a and c must satisfy the equation that

1/2*c*t^2 + (s-a)*t + 1 >= 0

In other words, the ant has a much better chance of catching the car at a very early time, before the car accelerates to infinite speeds.

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    $\begingroup$ No, $d_{ant}$ is not $at$. You forgot to take into account that when the elastic band stretches, it carries the ant along with it. $\endgroup$
    – Lopsy
    Nov 10, 2014 at 17:13
  • $\begingroup$ Good point. I was thinking of a string being unravelled. $\endgroup$
    – gopackstat
    Nov 10, 2014 at 19:47
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Ant reaches the car ONLY if car acceleration is low enough: a < [(πVa)^2+Vc^2]/2L

Where a= car acceleration, Va= ant speed, Vc=inital car speed, L=initial length of elastic.

This is more interesting problem than original 'Ant on Elastic' problem, since it is not given that ant will catch the car. But mathematical solution can follow similar principle ( calculate based on percentage position of ant on stick, instead of absolute position in meters ), with added complexity of acceleration.

Deriving formula:

  • if ant is standing, his position as percentage of elastic would be constant, so instead of solving with absolute ant position, we solve using ants "relative position" p ( p= antX(t)/L(t), p=0..1)
  • at any point in time 't', for some small additional time 'dt', ant would change his relative position for dp= Va * dt / L(t) = Va * dt / (L+a/2*t^2)
  • when 'p' reaches 1, ant has caught with the car. So : integral(dp) = 1 -> integral(t=0..T) (Va * dt / (L+a/2*t^2)) = 1
  • solving above result in T= tan(√(a*L/2Va^2)) / √(a/2L) ( time needed for ant to reach the car )
  • ant will catch the car when T is solvable from above formula. Since condition for tangent tan(x) is x< π/2, that means when √(a*L/2Va^2 < π/2
  • it expand to "ant will reach the car if a < (πVa)^2/2L"

BTW, if same is applied to problem with initial car speed Vc, then it result in:

  • solving integral(t=0..T) ( Va * dt / (L+ Vc * t + a/2*t^2) ) = 1
  • result in "ant will reach the car if a < [(πVa)^2+Vc^2]/2L"

I posted this (late) response since previous answers are not exactly right or complete:

  • most voted answer says that "ant never reaches the car, if the car is fast enough", but car speed (initial or at any other point) does not influence result, only time needed. It should have been "if the car acceleration is low enough". Also, formula provided there is missing dependence on initial elastic length, but can be ignored if "1" stated in the problem was 1m, and not eg 1cm or 1km
  • Next voted answer is "Yes", which is unconditional and incorrect.
  • Last answer (without any vote) is actually only one that was correct: "NO, if the car accelerates fast enough!" , but does not give formula for condition when and would reach the car.
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  • $\begingroup$ Where does this inequality condition come from? Can you edit to give a bit more explanation please? $\endgroup$
    – Rand al'Thor
    Aug 9, 2021 at 9:55
  • $\begingroup$ "Most importantly, ant reach car if acceleration is lower than some value" Yes, and that is already the case in the accepted answer. The ant's speed $a$ is greater than some expression involving $c$, the car's acceleration. If you rearrange it to isolate $c$, it will be less than some expression of $a$, and that is just your inequality. $\endgroup$
    – Jaap Scherphuis
    Aug 9, 2021 at 11:31
  • $\begingroup$ @jaap yes, it was use of 'a' for speed instead of acceleration that was an issue for me. As it is, accepted answer formula is ok ( missing dependence on initial elastic length, but can be ignored if it was 1m, and not eg 1cm or 1km ..) . Accepted answer itself is still mistaken in statement that ant never reach the car "if the car is fast enough" - as mentioned, speed of the car does NOT influence reachability ( neither initial speed not speed at any point). Speed of the car only influence time needed to reach the car. $\endgroup$
    – lost
    Aug 9, 2021 at 15:07
  • $\begingroup$ @lost I would say that "if the car is fast enough" is not wrong, just a bit ambiguous. I would interpret it as colloquial for "if the car is powerful enough", since the problem is all about the car's acceleration. Where the accepted answer lacks a bit is in the assumption that the car starts at rest. $\endgroup$
    – Jaap Scherphuis
    Aug 9, 2021 at 15:17
  • $\begingroup$ @rand I edited to expand on how formula was derived. It is possible to do same even with initial cars speed Vc, if at 2nd step you use dp= Va * dt / (L+a/2*t^2+Vc*t) $\endgroup$
    – lost
    Aug 9, 2021 at 15:59

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