Will the ant reach the car?

Question

A car starts distance 1 from a wall then drives away at constant speed $ 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 tension would slow the car).

Meanwhile, an ant crawls from the wall onto the elastic and then towards the car. On solid ground, the speed of the ant is $ a $. Needless to say, ants are slower than cars $ a \ll c $

Will the ant ever reach the car? Assume the elastic stretches without breaking.

Answer 1

This is the

Ant on a rubber rope problem.

The ant will eventually reach the car. The key to understand this is that the proportion between the total length of the elastic rope and the length walked by the ant can only increase with time, as it will eventually approach one.

Update

I'm sorry if my short solution makes this question seem too trivial. It is not, and in fact, the Wikipedia page I linked in the spoiler has a lengthy explanantion that covers both the mathematical aspect and the (un)intuitive approach to understand the problem better than I could explain here. Maybe an important detail not explicitly mentioned in the problem statement is that the ant has plenty of (unlimited) time to reach the car.

Answer 2

Let x(t) be the position of the ant at time t and l(t) the total length of the rubber band. Since the car is moving, the band will distort over time and this will affect the position of the ant. Assuming the elastic band distorts uniformly, we can say that :

• dl(t)/dt = c and l(t) = 1 + ct
• dx(t)/dt = a + x(t)/l(t) . dl(t)/dt
• dx(t)/dt = a + c.x(t)/(1 + ct)

We now have to solve the differential equation

y' + c/(1 + cx).y = a

The general solution of the homogeneous equation is

x_g(t) = k.(1 + ct)

Using the constant variation technique, we can find a particular solution of the differential equation :

x_h(t) = a(ct+1).ln(ct+1)/c

The overall solution is thus :

x(t) = x_s(t) + x_g = k(ct + 1) + a(ct+1).ln(ct+1)/c

We also have x(0) = 0, thus, x(0) = k = 0, and :

x(t) = a(ct+1).ln(ct+1)/c

Finally we can find the time t such that x(t) = l(t).
To do this, we must solve :

a.ln(ct+1) = c
ln(ct+1) = c/a
t = (e^c/a - 1)/c

Answer 3

Yes, the ant will reach the car. Calculus tells us exactly when:

Let C(t) be the position of the car at time t. The car travels at constant speed:

$ C(t) = 1 + ct $

Let A(t) be the position of the ant at time t. The speed of the ant is its speed on ordinary ground, plus a proportion of the car's speed (according to how far along the elastic it is):

$ \frac{dA}{dt} = a + { A \over { 1 + ct} } c = a + uc $

Making the natural substition $ u = {A \over {1+ct} } $ for the ants position as a proportion of the elastic. Then by the product rule

$ \frac{du}{dt} = { a \over {1+ct } } $

Which has solution

$ u = { a \over c } \log(1 + ct) $

Recalling that u is what proportion the ant is along the elastic, the ant reaches the car when u = 1 at

$ t = { {e^{c / a} - 1 } \over c } $

Edit: fixed missing c in exponent

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I'd love to see an intuitive argument that the ant will reach the car without appealing to calculus.

Answer 4

I'll try to keep the math as concise as possible. If we measure time from the moment a bit before the start of the experiment when extrapolating backwards the distance from the car to the wall would be$~0$, then at time $t$ the rubber rope will be stretched by a factor proportional to$~t$. If one uses marking on the rope itself to measure (relative) progress of the ant, then this stretching causes the effective speed relative to the markings to be divided by that stretching factor, so to be of the form $\frac bt$ where$~b>0$ is a constant depending on the unscaled speed of the ant and the above factor of proportionality.

Now if $t_0$ is the (positive) starting time, the progress of the ant at time $t_1>t_0$ is given by $\int_{t_0}^{t_1}\frac bt\mathrm d t$. The basic fact that shows that the ant will ultimately arrive, no matter what the constants are, is that

any anti-derivative of the function $t\mapsto\frac bt$ on the positive real numbers is unbounded above.

Indeed any anti-derivative is $t\mapsto b\ln(t)+C$, and the logarithm is unbounded, though it advances very slowly (whence the ant needs a lot of time). Concretely the progress of the ant at time $t_1$ is given by $\int_{t_0}^{t_1}\frac bt\mathrm d t=b(\ln(t_1)-\ln(t_0))$, and this gets as large as one wants by increasing $t_1$.

What this also shows is that if the car were to accelerate slightly giving a stretch factor $t^{1.1}$ at time$~t$, then the arrival of the ant would no longer be assured. Indeed an anti-derivative of $t\mapsto \frac b{t^{1.1}}$ is $-10\frac b{t^{0.1}}+C$, which function is increasing, but bounded above (by the constant $C$) as $t\to+\infty$.

Answer 5

The intuitive case is roughly this.

Assume the ant begins at some point, $ P $, which is some fraction of the distance along the rope. The ant travels at $a$ and the car travels at $c$.

Now imagine a point, $P'$, which is $ \frac{a}{2c} $ percent of the rope's total length towards the car. The ant will clearly travel that distance, given the ant will travel in some period of time $t$ the distance $\frac{a}{c}t$.

This continues, indefinitely, covering points $P''$ etc., until the ant reaches far enough down the line such that the distance between $P^n$ and the end of the rope is less than $\frac{a}{2c}$. At that point the ant will reach the end of the rope, and begin walking on the car (and eventually off of it!). Note, depending on $c$ and $a$ this may be a very long time.

That is because, while the rope continues to stretch indefinitely, the amount each stretch increases the length of the rope decreases over time proportional to the fraction of rope the ant has covered. At 1%, the rope stretching 1km more means the way 'ahead' increases by 0.99km; however, at 99%, the rope stretching 1km more means only a 0.01 km increase in the ant's total length covered. Eventually the remaining distance is less than the amount the ant can cover.

This isn't quite a 'limit' - eventually the ant would in fact walk past the car, assuming it still had the property of keeping the rope's proportion behind it, which isn't practical in a real life example, but this isn't practical either - which is what confuses people. It's similar to the reason why Achilles eventually catches the tortoise when you're running away: you can keep halving the distance between the two indefinitely, true, but time isn't really paying attention to that, it's just going on its merry way. (Of course, that paradox is a bit different, but in both cases the answer is that it does overtake despite the apparent limit problem.)

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I find it helpful to think of radial geometry here, as well. You can imagine traversing a circle of 360° at constant radius 1km, it will total 2$\pi$km distance covered to traverse. As long as your speed is positive, you will certainly traverse the entire circle.

Now, turn the circle into a spiral (say, a logarithmic spiral). Traversing a spiral starting at 1km radius, say the point $(x,y)=(1,0)$, and increasing in radius, you would always still eventually traverse that 360°, back to some point $(x,y)=(x,0)$, regardless of the speed at which you travel. Exactly the same problem as the circle, except you're making some points further 'out' than others - but still the same 360 degrees to travel, always traveling along, increasing what degree you're at (at least by a little!).

The second graph here gives a good example of how this will eventually happen. The parameters in the logarithmic spiral below can be manipulated to mimic the speed of the ant and the car. While it clearly may end up being a very long spiral, you can precompute the entire spiral given the relative speed of the car and the ant, and the starting length of the rope. It is fairly clear that no matter what that total spiral length is, given it is finite, the ant must eventually traverse all of it.

(sourced from Wolfram Alpha, searching for "logarithmic spiral")

Answer 6

An intuitive answer would be that the rubber band acts like an accelerating moving walkway for the ant. Every step it takes it's speed increases a little bit. At the start of the rubber band it's at a, and at the end it's at speed c+a. So it must be accelerating.

Since the ant is accelerating and the car's speed is constant. The ant will eventually catch the car.

Answer 7

a=speed of ant
c=speed of car(in units of distance per unit of time) we'll say feet and seconds.
t=time
l=length of band at time 0
l+ct=length of band at time t call it L
p=distance of ant from wall

c/L=Rate at which the band is growing in relation to its current size at any time. (also the rate in proportion to its current position at which any point on the band is increasing it's distance from the wall)
(L)
p*c/(l+ct)=how much the stretching of the band increases the ants distance from the wall at any time

this leaves you with the differential equation

$dp/dt=a+p*c/(l+ct)$

solve it for p given p(0)=0 and set the result = l+ct the solve for t to find how log it takes for and to reach end of band...I think. Please respond w/help/comments.

Answer 8

I will try for the intuitive argument When the problem starts the difference in speeds is c-a When the ant catches up to to the car and is crawling on the bumper it is c+a

the ant will be at the speed of the car at (1-a/c)*100% of the band. After that he is going faster and will eventually overtake the car

So can he get there? Well that point is no different then the car. He will be faster then it when he reaches (1-a/c)*100% to that point

And more importantly he will do it faster because it is closer to him

So now we can imagine that infinity until the ant's first step reaches this magic point of outrunning the next point. Another way of saying is that the velocity of one of these points will be less than a if the subdivision is preformed enough times.

As an example if the ant is going .5 c, he will reach the magic point at .5 of the band. The point to reach that is at .25, the point to accelerate over that spot is .125. And so towards 0. At some moment the ant's step will be greater than 1/(x^2)