【答案】
分析:(1)因?yàn)橹本€l的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/0.png)
x-3,并且與x軸、y軸分別相交于點(diǎn)A、B,所以分別令y=0;x=0,即可求出A、B的坐標(biāo);
(2)可設(shè)動(dòng)圓的圓心在C處時(shí)與直線l相切,設(shè)切點(diǎn)為D,連接CD,則CD⊥AD,CD=1,由∠CAD=∠BAO,∠CDA=∠BOA=90°,可知Rt△ACD∽R(shí)t△ABO,利用相似三角形對(duì)應(yīng)邊的比等于相似比,可得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/1.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/2.png)
,求出AC的值,即可得到此時(shí)OC的值,利用OC的長(zhǎng)度結(jié)合速度即可求出時(shí)間;根據(jù)對(duì)稱性,圓C還可能在直線l的右側(cè),與直線l相切,
此時(shí)OC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/4.png)
;
(3)可設(shè)在t秒時(shí),動(dòng)圓的圓心在F點(diǎn)處,動(dòng)點(diǎn)在P處,此時(shí)OF=0.4t,BP=0.5t,F(xiàn)點(diǎn)的坐標(biāo)為(0.4t,0),連接PF.
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/5.png">,又
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/6.png)
,所以可得到
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/7.png)
,進(jìn)而可得到FP∥OB,PF⊥OA,所以P點(diǎn)的橫坐標(biāo)為0.4t,又結(jié)合P點(diǎn)在直線AB上,可得P點(diǎn)的縱坐標(biāo)為0.3t-3,因此可見:當(dāng)PF=1時(shí),P點(diǎn)在動(dòng)圓上,當(dāng)0≤PF<1時(shí),P點(diǎn)在動(dòng)圓內(nèi),而當(dāng)P=1時(shí),由對(duì)稱性可知,有兩種情況:①當(dāng)P點(diǎn)在x軸下方時(shí),PF=-(0.3t-3)=1,解之可得t的值,②當(dāng)P點(diǎn)在x軸上方時(shí),PF=0.3t-3=1,解之得t的另一個(gè)值,進(jìn)而可得到當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/8.png)
時(shí),0≤PF≤1,并且此時(shí)點(diǎn)P在動(dòng)圓的圓面上,所經(jīng)過(guò)的時(shí)間為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/9.png)
.
解答:解:(1)在y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/10.png)
x-3中,令x=0,得y=-3;令y=0,得x=4,故得A、B兩的坐標(biāo)為
A(4,0),B(0,-3). (2分)
(2)若動(dòng)圓的圓心在C處時(shí)與直線l相切,設(shè)切點(diǎn)為D,如圖所示.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/images11.png)
連接CD,則CD⊥AD. (3分)
由∠CAD=∠BAO,∠CDA=∠BOA=90°,可知Rt△ACD∽R(shí)t△ABO,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/11.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/12.png)
,則AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/13.png)
. (4分)
此時(shí)OC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/14.png)
(秒). (5分)
根據(jù)對(duì)稱性,圓C還可能在直線l的右側(cè),與直線l相切,
此時(shí)OC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/15.png)
. (7分)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/16.png)
(秒).
答:(略). (8分)
(3)設(shè)在t秒,動(dòng)圓的圓心在F點(diǎn)處,動(dòng)點(diǎn)在P處,此時(shí)OF=0.4t,BP=0.5t,F(xiàn)點(diǎn)的坐標(biāo)為(0.4t,0),連接PF,
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/17.png)
,又
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/images20.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/19.png)
,
∴FP∥OB,∴PF⊥OA(9分)
∴P點(diǎn)的橫坐標(biāo)為0.4t,
又∵P點(diǎn)在直線AB上,
∴P點(diǎn)的縱坐標(biāo)為0.3t-3,
可見:當(dāng)PF=1時(shí),P點(diǎn)在動(dòng)圓上,當(dāng)0≤PF<1時(shí),P點(diǎn)在動(dòng)圓內(nèi). (10分)
當(dāng)PF=1時(shí),由對(duì)稱性可知,有兩種情況:
①當(dāng)P點(diǎn)在x軸下方時(shí),PF=-(0.3t-3)=1,解之得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/20.png)
;
②當(dāng)P點(diǎn)在x軸上方時(shí),PF=0.3t-3=1,解之得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/21.png)
. (11分)
∴當(dāng)時(shí)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/22.png)
時(shí),0≤PF≤1,此時(shí)點(diǎn)P在動(dòng)圓的圓面上,所經(jīng)過(guò)的時(shí)間為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/23.png)
,
答:動(dòng)點(diǎn)在動(dòng)圓的圓面上共經(jīng)過(guò)了
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021232002232001615/SYS201310212320022320016006_DA/24.png)
秒. (12分)
點(diǎn)評(píng):本題是一道綜合性強(qiáng)的題目,解決這類問(wèn)題常用到分類討論、數(shù)形結(jié)合、方程和轉(zhuǎn)化等數(shù)學(xué)思想方法.