【答案】
分析:(1)由于A(8,0),B(0,6),∴OB=6,OA=8,AB=10.在前3秒內,點P在OB上,點Q在OA上,設經(jīng)過t秒,點P,Q位置如圖.則OP=6-2t,OQ=t.∴△OPQ的面積A=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/0.png)
OP•OQ=t(3-t),當t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/1.png)
時,S
max=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/2.png)
.
(2)在前10秒內,點P從B開始,經(jīng)過點O,點A,最后到達AB上,經(jīng)過的總路程為20;點Q從O開始,經(jīng)過點A,最后也到達AB上,經(jīng)過的總路程為10.其中P,Q兩點在某一位置重合,最小距離為0.設在某一位置重合,最小距離為0.
設經(jīng)過t秒,點Q被點P“追及”(兩點重合),則2t=t+6,∴t=6.∴在前10秒內,P,Q兩點的最小距離為0,點P,Q的相應坐標為(6,0).
(3)①設0≤t<3,則點P在OB上,點Q在OA上,OP=6-2t,OQ=t.若PQ∥AB,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/4.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/6.png)
,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/7.png)
.
此時,P(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/8.png)
),Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/9.png)
,0).(2分)
②設3≤t≤7,則點P,Q都在OA上,不存在PQ平行于△OAB一邊的情況.
③設7<t<8,則點P在AB上,點Q在OA上,AP=2t-14,AQ=8-t.若PQ∥OB,
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/11.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/13.png)
,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/14.png)
.
此時,P(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/16.png)
),Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/18.png)
).(2分)
④設8≤t≤12,則兩點P,Q都在AB上,不存在PQ平行于△OAB一邊的情況.
⑤設12<t<15,則點P在OB上、點Q在AB上,BP=2t-24,BQ=18-t.
若PQ∥OA,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/20.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/22.png)
,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/23.png)
.
此時,P(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/24.png)
),Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/25.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/26.png)
).(2分)
解答:解:(1)∵A(8,0),B(0,6),∴OB=6,OA=8,AB=10.
在前3秒內,點P在OB上,點Q在OA上,設經(jīng)過t秒,點P,Q位置如圖.
則OP=6-2t,OQ=t.∴△OPQ的面積A=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/27.png)
OP•OQ=t(3-t),(2分)
當t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/28.png)
時,S
max=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/29.png)
.(2分)
(2)在前10秒內,點P從B開始,經(jīng)過點O,點A,最后到達AB上,經(jīng)過的總路程為20;點Q從O開始,經(jīng)過點A,最后也到達AB上,經(jīng)過的總路程為10.其中P,Q兩點在某一位置重合,最小距離為0.
設在某一位置重合,最小距離為0.
設經(jīng)過t秒,點Q被點P“追及”(兩點重合),則2t=t+6,∴t=6.∴在前10秒內,P,Q兩點的最小距離為0,點P,Q的相應坐標為(6,0).(4分)
(3)①設0≤t<3,則點P在OB上,點Q在OA上,OP=6-2t,OQ=t.
若PQ∥AB,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/30.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/31.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/32.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/33.png)
,解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/34.png)
.
此時,P(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/35.png)
),Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/36.png)
,0).(2分)
②設3≤t≤7,則點P,Q都在OA上,不存在PQ平行于△OAB一邊的情況.
③設7<t<8,則點P在AB上,點Q在OA上,AP=2t-14,AQ=8-t.
若PQ∥OB,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/37.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/38.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/39.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/40.png)
,解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/41.png)
.
此時,P(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/42.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/43.png)
),Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/44.png)
,0).(2分)
④設8≤t≤12,則兩點P,Q都在AB上,不存在PQ平行于△OAB一邊的情況.
⑤設12<t<15,則點P在OB上、點Q在AB上,BP=2t-24,BQ=18-t.
若PQ∥OA,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/45.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/46.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/47.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/48.png)
,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/49.png)
.
此時,P(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/50.png)
),Q(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/51.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/52.png)
).(2分)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231655334845420/SYS201310212316553348454015_DA/images53.png)
點評:此題很復雜,把動點問題與實際相結合,有一定的難度,解答此題的關鍵是分別畫出t在不同階段Q的位置圖,結合相應的圖形解答.