Question

摩天轮上的乘客在竖直平面内沿圆周运动。某个摩天轮半径为 $20$ 英尺，并以每分钟一圈的恒定速度旋转。乘客从摩天轮最低点到达比最低点高 $10$ 英尺的位置，需要多少秒？

Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?

$5$

$6$

$7.5$

$10$

$15$

答案：D

知识点：圆特殊直角三角形比与比例

难度评级：1580

解答：

设圆心高度为 $20$ 英尺。记圆心为 $O$ 、最低点为 $A$ ，最低点高度为 $0$ ；乘客到达高度 $10$ 时，比最低点高 $10$ 英尺。

从圆心向乘客所在高度作水平线，会形成一个直角三角形，其中竖直直角边长为 $10$ ，斜边即半径，长为 $20$ 。直角边是斜边的一半，所以乘客所在半径与竖直向下方向成 $60^\circ$ 。

摩天轮在 $60$ 秒内转过 $360^\circ$ ，所以转过 $60^\circ$ 需要 $\dfrac{60}{360}\cdot 60 = 10$ 秒。

所以正确答案是 D。

Put the center $O$ at height $20.$ The bottom $A$ is at height $0,$ and the rider reaches height $10,$ which is $10$ feet below the center.

The horizontal from the center down to the rider's level forms a right triangle where the vertical leg is $10$ and the hypotenuse (the radius) is $20.$ That leg is half the hypotenuse, so the radius to the rider makes $60^\circ$ with the downward vertical.

The wheel turns $360^\circ$ in $60$ seconds, so turning $60^\circ$ takes $\dfrac{60}{360}\cdot 60 = 10$ seconds.

Thus, the correct answer is D.

2002 AMC 10B 第 24 题

答案：D

难度评级：1580

解答：

其他年份的第 24 题