Picture: http://postimg.org/image/hxjgmp5f9/
In the figure, a kid at B is 6m ahead of a dog at A. They are running in the same
direction with constant speeds along a straight road. When the kid runs to C, the
dog runs to B. When the kid runs to D, the dog runs to C, and so on. The speed of the dog is 3 times that of the kid. How far does the dog have to run before
overtaking the kid?
I have no idea to calculate this question as I don't know what is the question
talking about...
Please help, thank you!
F.5 Maths a.s. and g.s.
2015-02-09 1:38 am
回答 (3)
2015-02-09 1:52 am
✔ 最佳答案
required distance=6+6(1/3)+6(1/3)^2+6(1/3)^3+... *
=6/(1-(1/3)) **
=9m
* AS with 1st term 6 and common ratio 1/3
** GS sum to infinity
參考: me
2015-02-09 6:57 pm
教科書想你懂得分別出那類題目是屬於a.s.或g.s.的，所以要求你用。
可以用得著就證明你懂得分，多一種方法解決問題。
此題是：
當小狗跑 6 m，孩子會跑了 6*(1/3) m；
當小狗跑 6*(1/3) m，孩子會跑了 6*(1/3)^2 m；
當小狗跑 6*(1/3)^2 m，孩子會跑了 6*(1/3)^3 m；
...
所以小狗共跑了：
6＋6*(1/3)＋6*(1/3)^2＋6*(1/3)^3＋⋯⋯
＝6/(1－1/3)
＝9 (m)
[這是 g.s., a＝6, r＝1/3, |r| < 1, sum to infinity＝a/(1－r)]
2015-02-09 10:59:36 補充：
此題只有g.s.用得著，不關a.s.事。
2015-02-10 11:20:30 補充：
可惜教科書從來都不會說它想學生知道什麼 (其實老師版的教科書有說明的)，
所以只能靠老師，老師不懂 (或不說)，學生也只能靠天份了。
可以用得著就證明你懂得分，多一種方法解決問題。
此題是：
當小狗跑 6 m，孩子會跑了 6*(1/3) m；
當小狗跑 6*(1/3) m，孩子會跑了 6*(1/3)^2 m；
當小狗跑 6*(1/3)^2 m，孩子會跑了 6*(1/3)^3 m；
...
所以小狗共跑了：
6＋6*(1/3)＋6*(1/3)^2＋6*(1/3)^3＋⋯⋯
＝6/(1－1/3)
＝9 (m)
[這是 g.s., a＝6, r＝1/3, |r| < 1, sum to infinity＝a/(1－r)]
2015-02-09 10:59:36 補充：
此題只有g.s.用得著，不關a.s.事。
2015-02-10 11:20:30 補充：
可惜教科書從來都不會說它想學生知道什麼 (其實老師版的教科書有說明的)，
所以只能靠老師，老師不懂 (或不說)，學生也只能靠天份了。
2015-02-09 4:43 am
我欣賞 fff 網友可以利用 G.S. sum to infinity 的概念作答。
讓我移到意見欄：
其實不理會 F.5 的 A.S. 和 G.S. 也可以用初中的想法計算。
Let the speed of the kid is x m/s, then the speed of the dog is 3x m/s.
Let t seconds be the time that the dog has to run before overtaking the kid.
Recall: Distance = Speed * Time
2015-02-08 20:43:54 補充：
Then, the distance traveled by the kid is xt m,
and the distance traveled by the dog is 3xt m,
Therefore,
3xt - xt = 6
2xt = 6
xt = 3
3xt = 9
The required distance traveled by the dog is 9 m.
2015-02-10 07:16:29 補充：
Re: YTC:
加油！
Re: 少年時 老師:
其實這題題目初次想不是太易。
因為小狗始終會追過孩子，但原來追過之前可以考慮用無限項之和．．．
讓我移到意見欄：
其實不理會 F.5 的 A.S. 和 G.S. 也可以用初中的想法計算。
Let the speed of the kid is x m/s, then the speed of the dog is 3x m/s.
Let t seconds be the time that the dog has to run before overtaking the kid.
Recall: Distance = Speed * Time
2015-02-08 20:43:54 補充：
Then, the distance traveled by the kid is xt m,
and the distance traveled by the dog is 3xt m,
Therefore,
3xt - xt = 6
2xt = 6
xt = 3
3xt = 9
The required distance traveled by the dog is 9 m.
2015-02-10 07:16:29 補充：
Re: YTC:
加油！
Re: 少年時 老師:
其實這題題目初次想不是太易。
因為小狗始終會追過孩子，但原來追過之前可以考慮用無限項之和．．．
收錄日期: 2021-04-15 18:13:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150208000051KK00085