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數學問題求詳解

Jul 19th 2014, 00:12

設甲速度為 u，乙速度為 v，跑道一個圈子長 d。
則甲第一次從背後追上乙需時 d/(u－v)，兩人再次相遇需時 d/(u＋v)。
此時乙恰好跑了四圈，所以，
4d＝v[d/(u＋v)＋d/(u－v)]
==> 4d(u＋v)(u－v)＝dv(u－v)＋dv(u＋v)
==> 4u^2－4v^2＝uv－v^2＋uv＋v^2
==> 4u^2－2uv^2－4v^2＝0
==> u＝v(1＋√17)/4 或 u＝v(1－√17)/4 (不合，負值)

答：甲的速度是乙的 (1＋√17)/4 倍

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