A $10 \mathrm{~m}$ long rubber cord has its $B$ end anchored to the ground, while the other end, $A$, is where a snail and a mischievous monkey are located. The snail starts moving towards $B$ at a speed of $1 \mathrm{~cm} / \mathrm{s}$, while the monkey pulls the end of the cord in the opposite direction at a speed of $10 \mathrm{~m} / \mathrm{s}$. Will the snail ever reach point $B$?
I. solution. At the end of the first second, the rope will be $2000 \mathrm{~cm}$ long, and during this time the snail will have moved $1 \mathrm{~cm}$. Therefore, after 1 second, the snail will be at least $1 \mathrm{~cm}$ away from the starting point. Thus, by the end of the first second, the snail will have covered at least $1 / 2000$ of the $10 \mathrm{~m}$. Similarly: at the end of the $k$-th second, the rope's length will be $(k+1) \cdot 1000 \mathrm{~cm}$; and the snail will be at least $1 \mathrm{~cm}$ away from the point where it was at the beginning of the second, so in the $k$-th second, the snail will cover at least $1 /(k+1)$ thousandth of the $10 \mathrm{~m}$. Therefore, by the end of the $n$-th second, the snail will have covered at least $1 / 2000 + 1 / 3000 + \ldots + 1 /(n+1)$ thousandth of the rope. Thus, if

$$
1 / 2 + 1 / 3 + \ldots + 1 /(n+1) > 1000
$$

for some suitable $n$, then the snail will reach $B$ despite the evil monkey's interference. The (1) is true because, as is known, for any large number $K$, there exists an $n$ such that

$$
1 + 1 / 2 + 1 / 3 + \ldots + 1 / n
$$

is greater than $K$.

II. solution. Let $S(t) \mathrm{~cm}$ be the distance of the snail from $B$ after $t$ time. Clearly, $S(0) = 1000$, and when the snail reaches the end point $B$ of the rubber rope, $S(t^*) = 0$. Let $v(t)$ be the snail's speed relative to point $B$. Clearly, $v(0) = 999$, because the monkey's speed is $1000 \mathrm{~cm} / \mathrm{s}$, and the snail's speed is in the opposite direction at $1 \mathrm{~cm} / \mathrm{s}$.

In general, the snail's speed at time $t$ is given by the following equality:

$$
v(t) = \frac{S(t)}{(t+1) 1000} 1000 - 1 = \frac{S(t)}{t+1} - 1
$$

Indeed, at time $t$, the rope's length is $(t+1) 1000$, and the snail is $S(t)$ away from $B$, so assuming the rope stretches uniformly, the point where the snail is located moves away from $B$ at a speed of $\frac{S(t)}{(t+1) 1000} 1000 \mathrm{~cm} / \mathrm{s}$ due to the monkey's speed of $1000 \mathrm{~cm} / \mathrm{s}$, and the snail moves towards $B$ at a speed of $1 \mathrm{~cm} / \mathrm{s}$ relative to the rope. Let $a(t)$ be the snail's acceleration after $t$ time. Using the relationships known from the physics of motion, since $\dot{v}(t) = a(t), \dot{S}(t) = v(t)$, by differentiating (2) we get

$$
a(t) = \frac{v(t)(t+1) - S(t)}{(t+1)^2} = \frac{S(t) - (t+1) - S(t)}{(t+1)^2} = -\frac{1}{t+1}
$$

Thus, the acceleration is available as a function of time, from which we can determine the speed by integration: $v(t) = -\ln (t+1) + c$. Where the constant $c$, based on the condition $v(0) = 999$, is 999, so

$$
v(t) = -\ln (t+1) + 999
$$

From this, by rearranging (2), we get that

$$
S(t) = (t+1)(-\ln (t+1) + 1000)
$$

We can see that for the value $t^* = e^{1000} - 1$, $S(t^*) = 0$, that is, the snail will reach $B$ after $e^{1000} - 1 \approx 1.97 \cdot 10^{434}$ seconds.