[解題] 國二理化直線運動
1.年級:國中二年級
2.科目:理化
3.章節:直線運動
4.題目:某物體初始速度為3 m/s,做等加速度運動前進,經一段時間後末速度為21 m/s,
總路徑長為S,試問當物體在1/2S位置時其速度為何?
5.想法:1.設在1/2S位置時速度為X m/s
2.先畫出V-t圖 → 將線段延長 → 可看到三個相似三角形
面積由小至大標為△1、△2、△3
(邊長比為3:X:21,面積比為9:X^2:441)
http://images.plurk.com/3289575_76504a3fd1dfcab7608464ee1831e61b.jpg
![](http://images.plurk.com/3289575_76504a3fd1dfcab7608464ee1831e61b.jpg)
3.△3 - △2 = △2 -△1
△3 + △1 = 2△2 --------(1)
4.△1:△2:△3 = 3^2: X^2 :21^2
可得△1 = (3/X)^2˙△2
△3 = (21/X)^2˙△2
將此結果帶入(1)
5.[(9 + 441)/X^2]˙△2 = 2△2
225 = X^2
X = 15
想請教各位是否有更快速的算法?
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