The Indianapolis F(t)

Can the Hare Beat the Tortoise? Given his ability to outpace Tortoise by $25\%$, the mathematically minded Hare wants to minimize his margin of victory over his longtime foe. The magical racetrack expands proportionally by 10 miles instantaneously at each minute. Based on the magically expanding track length, Hare wants to know:
How long after the race has begun should Hare wait so that both Tortoise and Hare will cross the finish line at the same exact moment?
First, we will figure out how long it will take Tortoise to finish. To do so, since the total length is dynamic but expands proportionally, we will instead focus on the ratio of the track completed at time $t$. The total track length is $$F(t) = 10 (1 + \lfloor t \rfloor).$$ Thus, while the Tortoise's speed may be fixed at $60$ mph with respect to a fixed vantage point, Tortoise's closing speed with respect to the track length is $$v_T(t) = \frac{ 1 \,\,\text{miles / minute} }{ F(t)\,\, \text{miles / track}} = \frac{1}{10 (1 + \lfloor t \rfloor)}\,\, \text{track / minute}.$$ Tortoise will then finish the track at time $\tau$ such that $$\int_0^\tau v_T(t) \,dt = \sum_{k=0}^{\lfloor \tau \rfloor} \frac{1}{10(k+1)} + \frac{\tau - \lfloor \tau \rfloor}{10 (1 + \lceil\tau \rceil)} = 1.$$ Ignoring the ceilings and floors, gives us roughly $\frac{1}{10} H_{\tau+1} \approx \frac{ \ln (\tau+1) + \gamma }{10} = 1,$ so $\tau \approx \lceil e^{10 - \gamma} \rceil - 1 = 12366$ where $\gamma=0.57721....$ is the Euler-Mascheroni constant. Root solving further gives $\tau = 12365.4681....$

Knowing when Tortoise will complete the track allows our very mathematically inclined Hare to back into when he should start. Hare's relative velocity is $$v_H(t) = \frac{3}{2(1+\lfloor t\rfloor)},$$ if $t \geq t_0$ and $v_H(t) = 0$ if $t \leq t_0.$ So then we need to find $t_0$ such that \begin{align*}\int_0^\tau v_H(t) \, dt &= \int_{t_0}^\tau v_H(t) \,dt \\&= \frac{ 3(\lceil t_0 \rceil - t_0) }{20 (1 + \lfloor t_0 \rfloor)} + \sum_{k=\lceil t_0 \rceil}^{\lfloor \tau \rfloor} \frac{3}{20(k+1)} + \frac{3(\tau - \lfloor \tau \rfloor)}{20(1 + \lceil \tau \rceil)} = 1.\end{align*} If we ignore the first and last terms then we have approximately $\frac{3}{20} \ln \frac{\tau}{t_0} \approx 1,$ which should give $t_0 \approx \tau e^{-20/3} \approx 15.$ Root solving further gives $$\mathbf{t_0 = 15.2416....} \,\, \textbf{minutes}.$$