Question

Michael 在一条长直路上以每秒 $5$ 英尺的速度步行。路上每隔 $200$ 英尺有一个垃圾桶。一辆垃圾车以每秒 $10$ 英尺的速度沿同一方向行驶，并在每个垃圾桶处停 $30$ 秒。当 Michael 经过一个垃圾桶时，他注意到前方的垃圾车刚离开下一个垃圾桶。Michael 和垃圾车会相遇多少次？

Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck travels at $10$ feet per second in the same direction as Michael and stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?

$4$

$5$

$6$

$7$

$8$

答案：B

知识点：相对速度路程、速度与时间过程模拟

难度评级：1860

解答：

给垃圾桶编号。在时刻 $0$ ，Michael 在 $0$ 号桶，垃圾车在 $1$ 号桶。Michael 在 $40n$ 秒时到达 $n$ 号桶。垃圾车在两个桶之间行驶 $20$ 秒，并停 $30$ 秒，所以它在 $50(n - 1)$ 秒离开 $n$ 号桶，并且（当 $n \ge 2$ 时）在 $50(n - 1) - 30$ 秒到达。

Michael 在 $n$ 号桶时垃圾车也在那里，恰好当 $50(n-1) - 30 \le 40n$ $\le 50(n-1)$ ，化简得 $5 \le n \le 8$ 。所以他们在 $5$ 号桶（ $t = 200$ ，垃圾车离开时）、 $6$ 号桶（ $t = 240$ ）、 $7$ 号桶（ $t = 280$ ）和 $8$ 号桶（ $t = 320$ ，垃圾车到达时）相遇。

在 $6$ 号桶和 $7$ 号桶之间，垃圾车（以 $10$ 英尺/秒移动）先超过 Michael，随后又被 Michael 追上一次，额外增加一次交会。总共相遇 $5$ 次。

因此，正确答案是 B。

Number the pails so Michael is at pail $0$ and the truck at pail $1$ at time $0.$ Michael reaches pail $n$ at $40n$ seconds. The truck spends $20$ seconds between pails and $30$ stopped, so it leaves pail $n$ at $50(n - 1)$ seconds and (for $n \ge 2$ ) arrives at $50(n - 1) - 30.$

Michael is at pail $n$ while the truck is there exactly when $50(n-1) - 30 \le 40n$ $\le 50(n-1),$ which simplifies to $5 \le n \le 8.$ So they meet at pail $5$ (at $t = 200,$ as the truck departs), pail $6$ ( $t = 240$ ), pail $7$ ( $t = 280$ ), and pail $8$ ( $t = 320,$ as the truck arrives).

Between pails $6$ and $7$ the truck (moving at $10$ ft/s) pulls ahead of and is then overtaken by Michael once more, adding one crossing. In all, they meet $5$ times.

Thus, the correct answer is B.

2008 AMC 12B 第 20 题

答案：B

难度评级：1860

解答：

其他年份的第 20 题