10.1 Move a particle, and when it travels the second mile and subsequent miles, its speed is inversely proportional to the number of miles traveled. And in each subsequent mile, the speed is constant. If it takes 2 hours to travel the second mile, then the time required to travel the $n$-th mile (in hours) is
(A) $\frac{2}{n-1}$.
(B) $\frac{n-1}{2}$.
(C) $\frac{2}{n}$.
(D) $2 n$.
(E) $2(n-1)$.
(20th American High School Mathematics Examination, 1969)
[Solution] According to the problem, the particle's velocity is a piecewise constant value. Let $v_{n}$ be the velocity at the $n$-th mile, and $t_{n}$ be the required time, then
$$
v_{n}=\frac{1}{t_{n}} .
$$

Also, $v_{n}$ is inversely proportional to $n-1$, so $t_{n}$ is directly proportional to $n-1$.
Thus, $t_{n}=k(n-1)$, where $k$ is to be determined.
When $n=2$, $t_{2}=k(2-1)=2$, so $k=2$.
$$
\therefore \quad t_{n}=2(n-1) \text {. }
$$

Therefore, the answer is $(E)$.