【答案】
分析:(1)把AD,DC,BC它們的和求出來再除以速度每秒3個單位就可以求出t的值;
(2)當點P運動到AD時上時,根據(jù)△APQ為直角三角形,△APQ的面積為S,點P和點Q的運動速度.即可求出S與t的函數(shù)關系式;同理,求出當點P運動到DC時上時的函數(shù)關系式.
(3)如圖,假設t秒后PQ∥DB,利用△PCN∽△PBQ,得出對應邊的比值,即可求出.
(4)假設存在t值,使PQ⊥AC,分四種情況討論即可.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/images0.png)
解:(1)如圖1,過C點作CE⊥AB,
∵直角梯形ABCD中,AB∥CD,∠A=90°,
∴四邊形ADCE是矩形,
∴AD=CE,AE=CD,
∵AB=6,AD=4,DC=3,
∴AD=CE=4,AE=CD=3,EB=AB-AE=3,
∴BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/0.png)
=5,
∴點P到達終點B時,所走的路程為AD+CD+BC=4+3+5=12,
∵點P從點A出發(fā)沿折線段AD-DC-CB以每秒3個單位長的速度向點B勻速運動,
∴當點P到達終點B時,t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/1.png)
=4.
答:t的值為4;
(2)當點P運動到AD時上時,
∵△APQ為直角三角形,△APQ的面積為S,
∴s=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/2.png)
PA•AQ,
∵點P從點A出發(fā)沿折線段AD-DC-CB以每秒3個單位長的速度向點B勻速運動,
點Q從點A出發(fā)沿射線AB方向以每秒2個單位長的速度勻速運動,
∴s=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/3.png)
×3t×2t=3t
2.
當點P運動到DC時上時,
s=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/4.png)
×AD×2t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/5.png)
×4×2t=4t,
答:點P運動到AD上時,S與t的函數(shù)關系式為s=3t
2;
當點P運動到DC時上時,S與t的函數(shù)關系式為s=4t,
(3)若PQ∥DB,則點P、Q必在DB同側.
①當點Q在AB上,點P在AD上時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/images7.png)
∵AP:AQ=3t:2t=3:2,
而AD:AB=4:6=2:3,
∴AP:AQ≠AD:AB,則此情景下PQ不平行DB;
②因點Q沿射線AB運動,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/images8.png)
所以點Q在AB延長線上,點P在CB上時,即當3<t<4 時,PB=12-3t,PC=3t-7,BQ=2t-6.
若PQ∥DB,設直線PQ交DC與N,
∵DC∥AB,
∴△PCN∽△PBQ,
∴CN:BQ=PC:PB,則CN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/6.png)
;
又∵NQ∥DB,
∴CN:CD=CP:CB,
則CN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/7.png)
,
所以
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/9.png)
,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/10.png)
(符合題意).
綜上情景①、②所述,當t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/11.png)
時,PQ∥DB.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/images15.png)
(4)存在t=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/12.png)
,使PQ⊥AC.理由如下:
分四種情況討論:
①當0<t≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/13.png)
時,P在AD上,Q在AE上,設PQ與AC交于點O;
如圖,若PQ⊥AC,則△AOP∽△ADC,∴AP:AC=AO:AD,∴3t:5=AO:4,∴AO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/14.png)
t,
又若PQ⊥AC,則△QOA∽△ADC,∴OA:DC=AQ:AC,∴AO:3=2t:5,∴AO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/15.png)
t,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/16.png)
t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/17.png)
t,∴t=0,此解不符合題意,則此時PQ⊥AC不成立;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/images22.png)
②當
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/18.png)
<t≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/19.png)
時,P在DC上,Q在AB上,設PQ與AC交于點O;
如圖,若PQ⊥AC,則△COP∽△CDA,∴CP:AC=OC:CD,∴(7-3t):5=OC:3,∴OC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/20.png)
(7-3t),
又若PQ⊥AC,則△QOA∽△ADC,∴OA:DC=AQ:AC,∴AO:3=2t:5,∴AO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/21.png)
t,
∵OC+OA=AC,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/22.png)
(7-3t)+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/23.png)
t=5,∴t=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/24.png)
,此解不符合題意,則此時PQ⊥AC不成立;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/images30.png)
③當
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/25.png)
<t≤3時,P在CB上,Q在AB上;
如圖,顯然此時PQ不可能與AC垂直;
④當3<t≤4時,P在CB上,Q在AB的延長線上,設直線PQ與AC交于點O,過點P作PM⊥AB于M.
在△BPM中,PM=BP•sin∠PBM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/26.png)
BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/27.png)
(12-3t),MQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/28.png)
.
由△QAO∽△ACD,得AO:AQ=CD:AC=3:5.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/images35.png)
過點P作PN⊥OQ交AB于N.則PN=BP=12-3t,BN=2BM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/29.png)
BP,
NQ=BN+BQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/30.png)
BP+(2t-6)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/31.png)
.
由△QOA∽△QPN,得AO:AQ=PN:NQ,
即3:5=BP:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/32.png)
,
∴25BP=18BP+30t-90,
∴7BP=7(12-3t)=30t-90,
∴51t=174,
解得t=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/33.png)
=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/34.png)
,
綜上可知,當t=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021233306884952627/SYS201310212333068849526024_DA/35.png)
時,PQ⊥AC.
點評:此題綜合性很強,把圖形的變換放在梯形的背景中,利用直角梯形的性質結合已知條件探究圖形的變換,根據(jù)變換的圖形的性質求出運動時間.此題屬于難題.