1952 AMC 12 第 25 题

Question

一名爆破工将引信设定为

30

秒后爆炸。他以每秒

8

码的速度跑开。声音传播速度为每秒

1080

英尺。当爆破工听到爆炸声时，他大约跑了：

A powderman set a fuse for a blast to take place in

30

seconds. He ran away at a rate of

8

yards per second. Sound travels at the rate of

1080

feet per second. When the powderman heard the blast, he had run approximately:

200

码

200

yd.

352

码

352

yd.

300

码

300

yd.

245

码

245

yd.

512

码

512

yd.

答案：D

知识点：路程、速度与时间单位换算

难度评级：1560

小提示：

设从点燃引信到听见声音的总秒数为

T

Let

T

be the total number of seconds from lighting the fuse until the sound is heard

大提示：

到时刻

T

时，声音已传播

T-30

秒，其路程必须等于爆破工跑出的距离

At time

T

, the sound has traveled for

T-30

seconds and must cover the runner’s distance

解答：

设总时间为

T

秒。爆破工与爆炸地点相距

8T

码，即

24T

英尺。因为声音在

30

秒后才开始传播，所以

1080(T-30)=24T\text{。}

因此

T=\frac{1350}{44}

秒，他跑出的距离为

8T=\frac{2700}{11}\approx245.45\text{ 码}\text{。}

选项中与之近似的是

245

码，所以正确答案是 D。

Let

T

be the total time in seconds. The powderman is

8T

yards, or

24T

feet, from the blast. Because the sound starts after

30

seconds,

1080(T-30)=24T.

Thus

T=\frac{1350}{44}

seconds, and the distance he ran is

8T=\frac{2700}{11}\approx245.45\text{ yards}.

Approximately

245

yards is the listed value, so the correct answer is D.

1952 AMC 12 第 25 题

答案：D

难度评级：1560

小提示：

大提示：

解答：

其他年份的第 25 题