1968 AMC 12 第 25 题

Question

艾斯以恒定速度跑步，弗拉什的速度是他的

x

倍，其中

x\gt1

。弗拉什让艾斯领先

y

码，发出信号后两人同向出发。则弗拉什追上艾斯前必须跑的码数为：

Ace runs with constant speed and Flash runs

x

times as fast,

x\gt1.

Flash gives Ace a head start of

y

yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:

xy

\dfrac{y}{x+y}

\dfrac{xy}{x-1}

\dfrac{x+y}{x+1}

\dfrac{x+y}{x-1}

若艾斯的速度为

v

，则两人的相对速度为

(x-1)v

If Ace’s speed is

v,

the closing speed is

(x-1)v

用追赶时间乘以弗拉什的速度

xv

Multiply the catch-up time by Flash’s speed

xv

若艾斯的速度为

v

，弗拉什的速度为

xv

，所以两人的相对速度为

(x-1)v

。追赶时间为

\frac{y}{(x-1)v}

，在此期间弗拉什跑了

\frac{xv\cdot y}{(x-1)v}=\frac{xy}{x-1}

。

因此，正确答案是 C。

If Ace runs at speed

v,

Flash runs at

xv,

so their closing speed is

(x-1)v.

The catch-up time is

\frac{y}{(x-1)v},

during which Flash runs

\frac{xv\cdot y}{(x-1)v}=\frac{xy}{x-1}.

Therefore, the correct answer is C.

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