Mathematical Formulation:

This problem describes a geometric progression of distances and times. Let $v_H$ be the velocity of the hare and $v_T$ be the velocity of the tortoise.

Initially, the separation is $x_0 = 10.0 \text{ km}$.

• Step 1: Hare covers $x_0$ in time $t_1$. Tortoise moves $x_1$ in time $t_1$.
  $$ t_1 = \frac{x_0}{v_H}, \quad x_1 = v_T t_1 = x_0 \frac{v_T}{v_H} $$
• Step 2: Hare covers $x_1$ in time $t_2$. Tortoise moves $x_2$ in time $t_2$.
  $$ t_2 = \frac{x_1}{v_H}, \quad x_2 = v_T t_2 = x_1 \frac{v_T}{v_H} = x_0 \left(\frac{v_T}{v_H}\right)^2 $$

Let the ratio $r = \frac{v_T}{v_H}$. Then $x_n = x_0 r^n$ and $t_n = t_1 r^{n-1}$.

Using the Given Data:

We are given:

• $x_3 = 0.08 \text{ m}$
• $t_7 = 1.28 \times 10^{-7} \text{ s}$

From the geometric progression relations:

$$ x_3 = x_0 r^3 \implies r^3 = \frac{0.08}{10000} = 8 \times 10^{-6} $$ $$ r = (8 \times 10^{-6})^{1/3} = 2 \times 10^{-2} = 0.02 $$

So, velocity ratio $\frac{v_T}{v_H} = 0.02$.

Now for the time $t_7$:

$$ t_7 = t_1 r^6 \implies t_1 = \frac{t_7}{r^6} $$ $$ t_1 = \frac{1.28 \times 10^{-7}}{(0.02)^6} = \frac{1.28 \times 10^{-7}}{64 \times 10^{-12}} = \frac{1.28}{64} \times 10^5 = 0.02 \times 10^5 = 2000 \text{ s} $$

Calculating Velocities: $$ v_H = \frac{x_0}{t_1} = \frac{10000 \text{ m}}{2000 \text{ s}} = 5.00 \text{ m/s} $$ $$ v_T = r \cdot v_H = 0.02 \times 5.00 = 0.10 \text{ m/s} $$

Total Time to Catch Up:

This is the sum of the infinite geometric series of time intervals:

$$ T = t_1 + t_2 + t_3 + \dots = \frac{t_1}{1-r} $$ $$ T = \frac{2000}{1 – 0.02} = \frac{2000}{0.98} \approx 2040.8 \text{ s} $$

Approximating to integer precision: $t \approx 2041 \text{ s}$.