【答案】
分析:(1)首先求得C的坐標(biāo),則M的坐標(biāo)即可求得,利用待定系數(shù)法即可求得函數(shù)的解析式;
(2)首先求得直線(xiàn)BC的解析式,當(dāng)Q和點(diǎn)R之間的距離為8時(shí),PQ一定在C點(diǎn)的右側(cè),則根據(jù)Q和點(diǎn)R之間的距離為8,即可得到一個(gè)關(guān)于x的方程,求得x的值,即E點(diǎn)的橫坐標(biāo),則BE即可求得,從而求得時(shí)間t;
(3))△MDC是等腰三角形,且是鈍角三角形,∠DMC是鈍角,且P和Q同時(shí)分別到達(dá)D和C,因而△MPQ的頂點(diǎn)P,Q在CD上移動(dòng)時(shí),三角形的三個(gè)角都可能是直角;
(4)首先判斷當(dāng)當(dāng)2≤d≤7時(shí),P,Q都在線(xiàn)段CD上,即可列不等式組求解.
解答:解:(1)∵梯形ABCD中,AB∥CD,D的坐標(biāo)是(0,4),CD=10,
∴C的坐標(biāo)是(10,4),
∴M的坐標(biāo)是(5,0),
設(shè)拋物線(xiàn)的解析式是:y=a(x-5)
2,把(0,4)代入得:25a=4,解得:a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/0.png)
,
則拋物線(xiàn)的解析式是:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/1.png)
(x-5)
2;
(2)設(shè)直線(xiàn)BC的解析式是y=kx+b,根據(jù)題意得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/2.png)
,解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/3.png)
,則直線(xiàn)的解析式是:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/4.png)
x+12,
根據(jù)題意得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/5.png)
(x-5)
2-(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/6.png)
x+12)=8,解得:x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/7.png)
(x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/8.png)
<0,故舍去),
則x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/9.png)
.即OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/10.png)
,BE=OB-OE=15-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/12.png)
,則t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/14.png)
;
(3)△MDC是等腰三角形,且是鈍角三角形,∠DMC是鈍角,且P和Q同時(shí)分別到達(dá)D和C.
因而△MPQ的頂點(diǎn)P,Q在CD上移動(dòng)時(shí),三角形的三個(gè)角都可能是直角,成為直角三角形;
點(diǎn)Q到達(dá)點(diǎn)D停止,但點(diǎn)P還在運(yùn)動(dòng),還會(huì)出現(xiàn)一個(gè)直角三角形,故t的值有4個(gè);
(4)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/images15.png)
作CF⊥AB于F.
則BF=5,
在直角△AOD中,AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/16.png)
=5,
∵點(diǎn)P從點(diǎn)A出發(fā)以每秒5個(gè)單位的速度沿AD向點(diǎn)D運(yùn)動(dòng),點(diǎn)E從點(diǎn)B出發(fā)以每秒5個(gè)單位的速度沿BO運(yùn)動(dòng).
∴P從A到D,以及E由B到F,即Q到達(dá)C,都需要1秒.
∵CD=10>7,
∴當(dāng)2≤d≤7時(shí),P,Q都在線(xiàn)段CD上.
設(shè)經(jīng)過(guò)x秒,P、Q相遇,則3(x-1)+5(x-1)=10,解得:x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/17.png)
,
設(shè)經(jīng)過(guò)t秒,P、Q兩點(diǎn)之間的距離為d,且2≤d≤7,當(dāng)P、Q相遇以前時(shí):則PQ=10-3(t-1)-5(t-1)=18-8t,
則2≤18-8t≤7,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/18.png)
≤t≤2.
相遇以后,即t≥
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/19.png)
時(shí):PQ=3(t-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/20.png)
)+5(t-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/21.png)
)=8t-18,則2≤8t-18≤7,當(dāng)3(t-1)=7時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/22.png)
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/23.png)
≤t≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/24.png)
.
總之,t的取值范圍是:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/25.png)
≤t≤2或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/26.png)
≤t≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192948472569571/SYS201311011929484725695025_DA/27.png)
.
點(diǎn)評(píng):本題考查了待定系數(shù)法求二次函數(shù)解析式,以及方程與不等式組的應(yīng)用,正確判斷當(dāng)2≤d≤7時(shí),P,Q都在線(xiàn)段CD上是關(guān)鍵.