Imagine a change in the famous story of the hare and the tortoise. In this new story, when the hare

Q: Why does the hare cover distance $x_0$ in time $t_1$ instead of $x_0 + x_1$?

The problem states 'it overcomes this distance in time $t_1$'. 'This distance' refers to the initial gap $x_0$. The hare runs a distance $x_0$, and during that same time, the tortoise moves forward by $x_1$.

Q: Why do the distances form a geometric progression?

In each step, the hare covers the previous gap $x_{n-1}$ in time $t_n = \frac{x_{n-1}}{v_H}$. In that same time, the tortoise moves $x_n = v_T t_n = x_{n-1} \frac{v_T}{v_H}$. Thus, each new gap is $\frac{v_T}{v_H}$ times the previous gap.

Q: How do we know the total time is just $\frac{x_0}{v_H - v_T}$?

This is the concept of relative velocity. The hare is closing the initial gap of $x_0$ at a relative speed of $v_H - v_T$. The infinite sum of times $t_1 + t_2 + \dots$ will converge exactly to this value.

Q: What if the ratio $\frac{v_T}{v_H}$ was greater than 1?

If $\frac{v_T}{v_H} \ge 1$, the tortoise would be faster than or equal to the hare, meaning the hare would never catch up, and the distances $x_n$ would grow or remain constant instead of shrinking.

Q: Why did we convert $10.0 \text{ km}$ to meters?

Standard SI units are essential for consistency. Since $x_3$ is given in meters and $t_7$ in seconds, converting $x_0$ to $10000 \text{ m}$ ensures our calculated speeds are in $\text{m/s}$.

Question

Imagine a change in the famous story of the hare and the tortoise. In this new story, when the hare wakes up, he finds the tortoise $x_0 = 10.0 	ext{ km}$ ahead moving with a constant velocity. The hare not ready to give up starts running again with a constant velocity. In its effort to win, it overcomes this distance in time $t_1$, but during this time the tortoise crawls further a distance $x_1$, the hare overcomes $x_1$ in time $t_2$, but the tortoise in this time crawls further a distance $x_2$. This situation continues repeatedly. A monkey, who was the referee measures only distance $x_3 = 0.08 	ext{ m}$ and time $t_7 = 1.28 	imes 10^{-7} 	ext{ s}$. Assuming the hare and the tortoise as particles, find their speeds. How long after the hare wakes up, will it win?

Sigma Percentile · Accepted Answer

The classic fable of the tortoise and the hare gets a fascinating mathematical upgrade in this problem! Instead of a simple race, we are presented with an infinite sequence of catch-up intervals, reminiscent of Zeno's Paradox. Let's dive into the mechanics of this infinite chase and see how a few precise measurements from a monkey referee can unravel the entire mystery.

### The Infinite Chase

Imagine the scene: the hare wakes up from his nap, only to realize the tortoise is already a massive $x_0 = 10.0 	ext{ km}$ ahead! The hare, determined to win, sprints at a constant velocity $v_H$. The tortoise, unfazed, continues plodding along at a constant velocity $v_T$.

The hare's first goal is to cover that initial $10.0 	ext{ km}$ gap. Let's call the time it takes to do this $t_1$. So, we can write:
$$t_1 = \frac{x_0}{v_H}$$

But here is the catch—during this time $t_1$, the tortoise hasn't been standing still. It has moved forward a new distance, let's call it $x_1$. This new distance is simply the tortoise's speed multiplied by the time $t_1$:
$$x_1 = v_T t_1 = x_0 \left(\frac{v_T}{v_H}ight)$$

Now, the hare has a new gap to close: $x_1$. It takes time $t_2$ to cover this new gap, but in that time, the tortoise moves forward by $x_2$. This creates an infinite sequence of shrinking gaps!

### Uncovering the Pattern

Let's look closely at the relationship between these distances. Every time the hare covers a gap $x_{n-1}$, the tortoise creates a new gap $x_n$. 
The time taken for the $n$-th step is:
$$t_n = \frac{x_{n-1}}{v_H}$$

And the new distance the tortoise covers in that time is:
$$x_n = v_T t_n = x_{n-1} \left(\frac{v_T}{v_H}ight)$$

Do you see the pattern? Each new gap is just the previous gap multiplied by the ratio of their speeds, $\frac{v_T}{v_H}$. This means the distances form a **geometric progression**! We can express any distance $x_n$ directly in terms of the initial gap $x_0$:
$$x_n = x_0 \left(\frac{v_T}{v_H}ight)^n$$

Similarly, we can express any time interval $t_n$:
$$t_n = \frac{x_0}{v_H} \left(\frac{v_T}{v_H}ight)^{n-1}$$

### The Monkey's Measurements

This is where our monkey referee comes in. He managed to measure exactly two obscure values: the distance $x_3 = 0.08 	ext{ m}$ and the time $t_7 = 1.28 	imes 10^{-7} 	ext{ s}$.

First, let's use $x_3$ to find the ratio of their speeds. We know $x_0 = 10000 	ext{ m}$. Plugging these into our distance formula for $n=3$:
$$x_3 = x_0 \left(\frac{v_T}{v_H}ight)^3$$
$$0.08 = 10000 \left(\frac{v_T}{v_H}ight)^3$$
$$\left(\frac{v_T}{v_H}ight)^3 = \frac{0.08}{10000} = 8 	imes 10^{-6}$$

Taking the cube root of both sides, we get a beautiful, clean ratio:
$$\frac{v_T}{v_H} = 2 	imes 10^{-2} = 0.02$$

Now, let's use the monkey's second measurement, $t_7$, to find the actual speeds. Using our time formula for $n=7$:
$$t_7 = \frac{x_0}{v_H} \left(\frac{v_T}{v_H}ight)^6$$
$$1.28 	imes 10^{-7} = \frac{10000}{v_H} (0.02)^6$$

Let's calculate $(0.02)^6$. Since $0.02 = 2 	imes 10^{-2}$, raising it to the 6th power gives $64 	imes 10^{-12}$.
$$1.28 	imes 10^{-7} = \frac{10000 	imes 64 	imes 10^{-12}}{v_H}$$
$$1.28 	imes 10^{-7} = \frac{6.4 	imes 10^{-7}}{v_H}$$

Solving for $v_H$, we find the hare's speed:
$$v_H = \frac{6.4 	imes 10^{-7}}{1.28 	imes 10^{-7}} = \mathbf{5.00 	ext{ m/s}}$$

Since the tortoise's speed is $0.02$ times the hare's speed, we easily find:
$$v_T = 0.02 	imes 5.00 = \mathbf{0.10 	ext{ m/s}}$$

### The Final Catch

We have the speeds, but the ultimate question remains: how long does it take for the hare to finally catch the tortoise and win the race?

While we could sum the infinite geometric series of times ($t_1 + t_2 + t_3 + \dots$), there is a much more elegant way using **relative velocity**.

From the hare's perspective, the tortoise is moving towards him at a relative speed of $v_H - v_T$. The hare simply needs to close the initial gap of $x_0$ at this relative speed. 
$$t = \frac{x_0}{v_H - v_T}$$
$$t = \frac{10000}{5.00 - 0.10} = \frac{10000}{4.90}$$
$$t \approx \mathbf{2041 	ext{ s}}$$

And there you have it! By recognizing the geometric progression hidden within the chase, we transformed a seemingly impossible infinite sequence into a straightforward calculation. The hare wins, but the math is the real champion here!

Visualized Solution

$x_{0} = 10.0 km$

$t_{1} = \frac{x _{0}}{v _{H}}$

$x_{n} = x_{0} (\frac{v _{T}}{v _{H}})^{n}$

$\frac{v _{T}}{v _{H}} = 0.02$

$v_{H} = 5.00 m/s$

$v_{T} = 0.10 m/s$

$t \approx 2041 s$

The Sigma Insight: Relative Velocity

The Infinite Chase

Uncovering the Pattern

The Monkey's Measurements

The Final Catch

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