AMC12 2011 B

AMC12 2011 B · Q8

AMC12 2011 B · Q8. It mainly tests Rates (speed), Circle theorems.

Keiko walks once around a track at the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?

Keiko 每天以相同的恒定速度绕跑道走一圈。跑道两侧为直线，两端为半圆。跑道宽度为 $6$ 米，她绕外侧边缘走一圈比绕内侧边缘多花 $36$ 秒。Keiko 的速度是多少米每秒？

(A) \frac{\pi}{3} \frac{\pi}{3}

(B) \frac{2\pi}{3} \frac{2\pi}{3}

(D) \frac{4\pi}{3} \frac{4\pi}{3}

(E) \frac{5\pi}{3} \frac{5\pi}{3}

Answer

Correct choice: (A)

正确答案：（A）

Solution

To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in the time it takes her for each distance. We are given the difference in time, so all we need to find is the difference between the distances. The track is divided into lengths and curves. The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves. The curves of the track are semicircles, but since there are two of them, we can consider both of the at the same time by treating them as a single circle. We need to find the difference in the circumferences of the inside and outside edges of the circle. The formula for the circumference of a circle is $C = 2 * \pi * r$ where $r$ is the radius of the circle. Let's define the circumference of the inside circle as $C_1$ and the circumference of the outside circle as $C_2$. If the radius of the inside circle ($r_1$) is $n$, then given the thickness of the track is 6 meters, the radius of the outside circle ($r_2$) is $n + 6$. Using this, the difference in the circumferences is: $C_2 - C_1 = 2 * \pi * (r_2 - r_1) = 2 * \pi * (n + 6 - n) = 2 * \pi * 6 = 12\pi$ $12\pi$ is the difference between the inside and outside lengths of the track. Divided by the time differential, we get: $12\pi \div 36 = \boxed {\textbf{(A)}\ \frac{\pi}{3}}$

要找 Keiko 的速度，只需计算绕跑道内侧边缘与外侧边缘一圈的路程差，再除以所用时间差。题目已给出时间差，因此只需求路程差。跑道由直线段和弧线段组成。直线段在内外侧的长度相同，因此只需考虑弧线部分。两端是半圆，合起来可视为一个整圆。我们需要求内侧圆周与外侧圆周的周长差。圆周长公式为 $C = 2 * \pi * r$，其中 $r$ 为半径。设内侧圆周为 $C_1$，外侧圆周为 $C_2$。若内侧半径 $r_1$ 为 $n$，由于跑道宽度为 6 米，外侧半径 $r_2$ 为 $n+6$。则周长差为： $C_2 - C_1 = 2 * \pi * (r_2 - r_1) = 2 * \pi * (n + 6 - n) = 2 * \pi * 6 = 12\pi$ 因此内外侧一圈路程差为 $12\pi$。除以时间差 $36$ 秒，得速度： $12\pi \div 36 = \boxed {\textbf{(A)}\ \frac{\pi}{3}}$

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