2010 AMC 8 第 8 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

8.

Emily 沿一条笔直长路骑自行车时,看到 Emerson 在她前方 1/21/2 英里处同方向滑行。她超过他后,可以从后视镜里看到他,直到他落在她后方 1/21/2 英里处。Emily 以每小时 1212 英里的恒定速度骑行,Emerson 以每小时 88 英里的恒定速度滑行。Emily 能看到 Emerson 多少分钟?

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction 1/21/2 mile in front of her. After she passes him, she can see him in her rear view mirror until he is 1/21/2 mile behind her. Emily rides at a constant rate of 1212 miles per hour, and Emerson skates at a constant rate of 88 miles per hour. For how many minutes can Emily see Emerson?

 6\ 6

 8\ 8

 12\ 12

 15\ 15

 16\ 16

答案:D
知识点:相对速度路程、速度与时间单位换算
难度评级:1100
解答:

dd 为 Emily 领先 Emerson 的距离。Emily 能看到 Emerson 当且仅当 12d12-\dfrac 12 \leq d \leq \frac 12。其中 tt 以小时为单位;令 t=0t= 0d=12d = -\frac 12。根据两人的相对速度,d=(128)t12d = (12-8)t - \frac 12 。当 d=12d = \frac 12 时, 因此经过 0.250.25 小时,也就是 600.25=1560\cdot0.25 = 15 分钟。 12=4t12    1=4t\frac 12 = 4t - \frac 12 \implies 1 = 4t     t=0.25.\implies t = 0.25.

所以正确答案是 D

Let dd be how far Emily is ahead of Emerson. Emily sees Emerson if 12d12.-\dfrac 12 \leq d \leq \frac 12. Suppose at t=0,t= 0, where tt is in hours, that d=12.d = -\frac 12. Then, d=(128)t12.d = (12-8)t - \frac 12 . Since we must find where d=12,d = \frac 12, we find the time where 12=4t12    1=4t\frac 12 = 4t - \frac 12 \implies 1 = 4t     t=0.25.\implies t = 0.25. Since 0.250.25 hours passed, we know that 600.25=1560\cdot0.25 = 15 minutes passed.

Therefore, the answer is D .

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