[【學科】] 【物理】距離問題

33144653

下載 (29.18 KB)

2017-11-19 20:15

39475494

我之前也解了，我貼我解的

令 q/p = m
1+l1/x = (1+m)²
1+(l1+l2)/x = (1+2m)²
l1/x = (1+m)² - 1 = (2+m)m
(l1+l2)/x = (1+2m)² - 1 = (2+2m)2m
1+l2/l1 = (4+4m)/(2+m) = 4 - 4/(2+m)
m = 4/(3-l2/l1) - 2 = 4l1/(3l1-l2) - 2
l1/x = (2+m)m = [4l1/(3l1-l2)][4l1/(3l1-l2) - 2]
=4l1(2l2-2l1)/(3l1-l2)²
x = (3l1-l2)²/(8l2-8l1)

33144653

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2017-11-21 22:32

下載 (157.17 KB)

2017-11-21 22:33

詳解~

39475494

OA = x，時間 p
AB = l1，時間 q
BC = l2，時間 q
初速為 0 時(O為起點時)，S = at²/2
x = ap²/2
x+l1 = a(p+q)²/2
x+l1+l2 = a(p+2q)²/2

1+l1/x = (1+q/p)²
1+(l1+l2)/x = (1+2q/p)²
解 x 和 q/p 應該解的出來    ← l1 l2 是已知