Can You Race Against Zeno?

You run a $5{,}000$-metre race starting at a $24$-minute pace. When you reach the halfway point ($2{,}500$ m), you speed up by $10\%$. At the three-quarter point you speed up another $10\%$, and so on: every time you reach the midpoint between your current position and the finish, your speed jumps by $10\%$. How long does the race take?

The Fiddler, Zach Wissner-Gross, June 13, 2025(original post)

Solution

Let the base speed be $v_0$ (so $5000/v_0=24$ minutes). The $k$-th segment covers half of the remaining distance, $5000/2^{k+1}$, at speed $v_0(1.1)^k$, taking $$t_k=\frac{5000/2^{k+1}}{v_0(1.1)^k}=24\cdot\frac12\Bigl(\frac1{2.2}\Bigr)^{k}.$$ Summing the geometric series, $$T=\sum_{k=0}^{\infty}t_k=24\cdot\frac12\cdot\frac{1}{1-\tfrac1{2.2}} =24\cdot\frac{11}{12}=\boxed{22\text{ minutes}}.$$

The computation

Run the race rather than the series: at base pace $24/5000$ minutes per metre, repeatedly cover half the remaining distance at the current speed, add the time, then speed up by $10\%$. The total converges to $22$.

pace0 = 24 / 5000; remaining = 5000.0; mult = 1.0; T = 0.0
for _ in range(2000):                    # halving leaves the tail negligible
    seg = remaining / 2
    T += seg * pace0 / mult              # time for this segment at current speed
    remaining -= seg; mult *= 1.1        # reach the new midpoint, speed up 10%
print(round(T, 4))                       # 22.0

Extra Credit

Now the speed-up is continuous: wherever you are on the course, you run $10\%$ faster than you were when you were twice as far from the finish. The speed must rise smoothly. How long does the race take?

Solution

Write $v(x)$ for your speed when $x$ metres from the finish. The rule is $v(x)=1.1\,v(2x)$. A power law $v(x)=C x^{a}$ satisfies it when $(\tfrac12)^a=\tfrac1{1.1}$, i.e. $a=-\log_2 1.1$, and smoothness (a continuous, monotone speed) forces this single solution. With $v(5000)=5000/24$, $$T=\int_0^{5000}\frac{dx}{v(x)}=\frac{24}{1-a}=\frac{24}{1+\log_2 1.1}\approx\boxed{21.10\text{ minutes}}.$$ (The source’s value is paywalled; this closed form is my own.)

The computation

Fix $v(x)=Cx^{a}$ from the scaling rule and the finish-line pace, then numerically integrate the race time $\int_0^{5000} dx/v(x)$ rather than evaluating the closed form.

from math import log2
from scipy.integrate import quad
a = -log2(1.1)                           # v(x) = C x^a satisfies v(x) = 1.1 v(2x)
C = (5000 / 24) / 5000**a                # so that v(5000) = 5000/24
T, _ = quad(lambda x: x**(-a) / C, 0, 5000)
print(round(T, 4))                       # 21.0988