Ant on a Rubber Band

Answer: Yes, the ant reaches the racecar.

Part 1: Intuition

At first glance, this seems impossible. The car moves at 1 m/s while the ant crawls at just 1 cm/s—the car is 100 times faster. How could the ant ever catch up?

The key insight is to think about the ant’s position as a percentage of the rubber band, not its absolute position in metres.

The rubber band stretches uniformly along its entire length. If the ant stops crawling at the 25% mark, it stays at 25% even as the band stretches—every point on the band moves proportionally away from the wall, so the ant gets “carried along” by the stretching.

This means walking can only increase the ant’s percentage. In the first second, the ant walks 1 cm on a 1 m band—about 1% progress. In the second second, the band is 2 m long, so 1 cm is about 0.5% progress. In the third second, it’s about 0.33% progress. The gains get smaller, but they’re always positive. The ant’s percentage never decreases.

There’s another way to see why the ant eventually wins. The farther along the ant gets, the less the stretching hurts. When the ant is near the wall, a 1-metre stretch adds almost the full metre to the distance remaining. But when the ant is halfway across, that same stretch only adds 0.5 metres ahead—the other 0.5 metres stretches behind the ant. Eventually, the ant will be so far along that each 1-metre stretch adds less than 1 cm ahead, and the ant gains ground with every step.

So yes, the ant reaches the racecar.

Part 2: Understanding It Discretely

Let’s make this precise by thinking about the ant’s position as a fraction of the band’s length.

When the band stretches uniformly, every point moves proportionally away from the wall. If the ant is 10% of the way along, stretching doesn’t change that percentage—the ant gets “carried along.” The only thing that changes the ant’s fractional position is its own crawling.

To make things simpler, we'll assume the ant crawls forward 1cm first, then the car will move 1 metre forward. Working in 1-second steps:

• Second 1: The band is 1m long. The ant crawls 1cm, covering $\frac{1}{100}$1001​ of the band.
• Second 2: The band is 2m long. The ant crawls 1cm, covering $\frac{1}{200}$2001​ of the band.
• Second 3: The band is 3m long. The ant crawls 1cm, covering $\frac{1}{300}$3001​ of the band.
• Second $n$n: The band is $n$n metres long. The ant covers $\frac{1}{100n}$100n1​ of the band.

After $n$n seconds, the ant’s total fractional progress is:

$\frac{1}{100}+\frac{1}{200}+\frac{1}{300}+\cdots +\frac{1}{100n}=\frac{1}{100}(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n})$1001​+2001​+3001​+⋯+100n1​=1001​(1+21​+31​+⋯+n1​)

The sum in parentheses is the harmonic series, $H_n=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}$Hn​=1+21​+31​+⋯+n1​.

The harmonic series diverges—it grows without bound, albeit slowly (see The Harmonic Series).

Since $H_n\to \infty$Hn​→∞, eventually $H_n>100$Hn​>100, and the ant’s fractional position exceeds 1. The ant has reached the car.

Part 3: Computing the Exact Time

So far, we've been assuming discrete steps: the ant moves first, then the car moves. To find precisely when the ant arrives, we have to imagine both of them moving at the same time.

At time $t$t, the band has length $L\left(t\right)=1+t$L(t)=1+t metres. The ant crawls at 0.01 m/s, so its fractional progress rate is $\frac{0.01}{1+t}$1+t0.01​.

Let $p\left(T\right)$p(T) be the fractional progress along the rubber band at time $T$T, where 0 represents the start and 1 represents the end. Integrating from $t=0$t=0 to $t=T$t=T:

$p\left(T\right)={\int }_0^T\frac{0.01}{1+t}dt=0.01\ln (1+T)$p(T)=∫0T​1+t0.01​dt=0.01ln(1+T)

The ant reaches the car when $p\left(T\right)=1$p(T)=1:

Since $e^{100}\approx 2.69\times {10}^{43}$e100≈2.69×1043, this is roughly $8.5\times {10}^{35}$8.5×1035 years—vastly longer than the age of the universe.

But mathematically, the ant does reach the racecar.