解:(1)如圖1.設(shè)動點(diǎn)P出發(fā)t秒后,點(diǎn)P到達(dá)點(diǎn)A且點(diǎn)Q正好到達(dá)點(diǎn)C時,BC=BA=t,
則S
△BPQ=

BC•CD=

×t×8=40,
所以t=10(秒),
則BC=BA=10cm,點(diǎn)M的坐標(biāo)為(10,40).
過點(diǎn)A作AH⊥BC于H,則四邊形AHCD是矩形,
∴AD=CH,CD=AH=8cm,
在Rt△ABH中,∵∠AHB=90°,AB=10cm,AH=8cm,
∴BH=

=6cm,
∴CH=BC-BH=4cm,
∴AD=4cm;

(2)如備用圖1,延長AD到A′,使A′D=AD,連接A′B,交CD于P,
則PA+PB=PA′+PB=A′B最�。�
∵A′D∥BC,
∴△A′DP∽△BCP,
∴

=

,即

=

,
解得DP=

,
∴BA+AD+DP=10+4+

=

,
∴t=

÷1=

.
故P在CD邊上運(yùn)動時,存在時刻t=

秒,能夠使△PAB的周長最�。�
(3)△PCD為等腰三角形時,分三種情況:
①如果PC=PD,如備用圖2,作CD的垂直平分線交AB于P
1,則P
1為AB的中點(diǎn),此時t
1=BP
1÷1=5;

②如果CP=CD=8,如備用圖3,以C為圓心,CD長為半徑畫弧,交AB于點(diǎn)P
2,過P
2作P
2E⊥BC于E,過點(diǎn)A作AH⊥BC于H.
設(shè)BP
2=x,則P
2E=BP
2•sin∠B=x•

=

x,BE=BP
2•cos∠B=

x,
∴CE=BC-BE=10-

x.
在Rt△P
2EC中,∵∠P
2EC=90°,
∴P
2E
2+CE
2=CP
22,(

x)
2+(10-

x)
2=64,
整理,得x
2-12x+36=0,

解得x
1=x
2=6,
∴BP
2=6,t
2=BP
2÷1=6;
③如果DP=DC,如備用圖4,以D為圓心,CD長為半徑畫弧,交AB于點(diǎn)P
3,過P
3作P
3F⊥AD于F.
設(shè)AP
3=y,則P
3F=AP
3•sin∠FAP
3=AP
3•sin∠B=y•

=

y,AF=AP
3•cos∠B=

y,
∴DF=DA+AF=4+

y.
在Rt△P
3FD中,∵∠P
3FD=90°,
∴P
3F
2+DF
2=DP
32,(

y)
2+(4+

y)
2=64,
整理,得5y
2+24y-240=0,
解得y
1=

,y
2=

(不合題意舍去),
∴BP
3=AB-AP
3=10-

=

,t
3=BP
3÷1=

;
綜上所述,△PCD能成為等腰三角形,此時t的值為5秒或6秒或

秒;

(4)當(dāng)點(diǎn)P在BA邊上時,
y=

×t×tsinB=

t
2×

=

t
2(0≤t≤10);
當(dāng)點(diǎn)P在DC邊上時,
y=

×10×(22-t)=-5t+110(14≤t≤22);
如圖2所示.
分析:(1)P在AD邊上運(yùn)動時,△BPQ以BQ為底邊,以CD長為高,因此可根據(jù)△BPQ的面積為40cm
2求出BC=10cm,而P、Q速度相同,P到A的時間與Q到C的時間相同,因此BA=BC=10cm,點(diǎn)M的坐標(biāo)為(10,40).求AD的長可通過構(gòu)建直角三角形來求解.過A作AH⊥BC與H,那么在直角三角形ABH中,AH=CD=8cm,BA=10cm,因此可根據(jù)勾股定理求出BH=6cm,那么AD=BC-BH=4cm;
(2)△PAB的周長=PA+AB+PB,而AB=10cm為定值,所以當(dāng)PA+PB最小時,△PAB的周長最小.延長AD到A′,使A′D=AD,連接A′B,交CD于P,此時PA+PB最小.由△A′DP∽△BCP,根據(jù)相似三角形對應(yīng)邊成比例即可求解;
(3)△PCD為等腰三角形時,分三種情況進(jìn)行討論:①PC=PD;②CP=CD;③DP=DC;
(4)△BQP中,BQ=t,BP=t,以BQ為底邊的高可用BP•sin∠B來表示,然后可根據(jù)三角形的面積計(jì)算公式得出關(guān)于y,t的函數(shù)關(guān)系式.
點(diǎn)評:本題是四邊形綜合題,主要考查了梯形的性質(zhì),三角形的面積,解直角三角形,相似三角形的判定與性質(zhì),軸對稱-最短路線,等腰三角形的性質(zhì)等知識,綜合性較強(qiáng),有一定難度.借助函數(shù)圖象表達(dá)題目中的信息,讀懂圖象是關(guān)鍵.