矩形ABCD中，AB=6，AD=6，將矩形ABCD在直線l上按順時針方向不滑動地每秒轉90*，轉動3秒後，頂點A經過的路線長是?
D____________C
A____________B
AD,BC are linked.
Mathematical Problem
2007-08-02 7:45 pm
回答 (3)
2007-08-04 2:26 am
✔ 最佳答案
在第1秒，A 的路線等於以AB 為半徑來畫的圓周的1/4，即: 2 x 6 x 丌 x 1/4 = 3丌
第2秒，A 的路線等於以AC 為半徑來畫的圓周的1/4，
即: 2 x sqrt(6^2 + 6^2) x 丌 x 1/4 = 3sqrt(2) 丌
第3秒，A 的路線等於以AD 為半徑來畫的圓周的1/4，
即: 2 x 6 x 丌 x 1/4 = 3丌
所以A 經過的路線長
= 3丌 + 3sqrt(2) 丌 + 3丌
=[ 6 + 3sqrt(2) ] 丌
= 32.2 (取值至三位有效數字)
2007-08-03 4:21 am
Ans = (2*pr*6)/4+(2*pr*(sq root of(6<sup>2</sup>+6<sup>2</sup>))/4+(2*pr*6)/4
2007-08-02 20:29:11 補充：
Ans = 3*pr + 4.24*pr + 3*pr = 10.24*pr = 32.1536
2007-08-02 20:29:11 補充：
Ans = 3*pr + 4.24*pr + 3*pr = 10.24*pr = 32.1536
2007-08-02 8:30 pm
If you rotate it for a cycle, you will see its path is a circle.
A period need 4s, so in 3s, A has moved for 270 degree or 3/4 circle
Distance = 3/4 of the circle = 3/4 (2r pi)
r=(6^2+6^2)^0.5=6(2)^0.5. pi=3.14159
Distance= 3/2 r pi =39.99 cm
Displacement = 6 cm (A now is at B)
A period need 4s, so in 3s, A has moved for 270 degree or 3/4 circle
Distance = 3/4 of the circle = 3/4 (2r pi)
r=(6^2+6^2)^0.5=6(2)^0.5. pi=3.14159
Distance= 3/2 r pi =39.99 cm
Displacement = 6 cm (A now is at B)
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