## 235. Math Puzzle 12/84

The problem statement can be found in the article "Thinking is a matter of luck - or?" [in the technikusHeft 12/84] on pages 40/41.

Calculate, therefore, after how many seconds and also meters Achilles will have really caught up with the tortoise.

We assume the speed of Achilles to be $6 \mathrm{~m} / \mathrm{s}$ and that of the racing tortoise to be $0.5 \mathrm{~m} / \mathrm{s}$. Also consider the real head start of the reptile, which is $184.97 \mathrm{~m}$.
$v=s / t ; v_{A}=1, v_{S K}=\frac{1}{12} ;$ Lead of SK $=n$

When they meet, they are equally far from the start:

$$
s_{A}=1 \cdot t=\frac{1}{12} t+n=s_{S K} \quad, \quad t=\frac{12}{11} n=n \cdot 1.0909 \ldots
$$

Example:

$v_{A}=6 \mathrm{~m} / \mathrm{s} ; v_{S K}=0.5 \mathrm{~m} / \mathrm{s} ; n=184.97,184.97 \mathrm{~m} \cdot 1.0909=201.78 \mathrm{~m}$

Time for Achilles to cover 1 stadium: $t_{A}=\frac{186.97}{6}=30.83 \mathrm{~s}$

$30.83 \mathrm{~s} \cdot 1.0909=33.6 \mathrm{~s}$. They meet after $33.6 \mathrm{~s}$ at $201.78 \mathrm{~m}$.