【答案】
分析:(1)已知了AB的長和B點的坐標,那么sin∠BAO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/1.png)
,因此∠BAO=60°
(2)由函數(shù)的圖形可知:當(dāng)t=5時,三角形OPQ的面積是30,如果設(shè)點P的速度為a,那么AP=5a,那么P到AC的距離就是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/2.png)
a,也就是P到OQ的距離為10-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/3.png)
a.OQ=QD+OD=5a+2.因此(5a+2)×(10-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/4.png)
)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/5.png)
=30,解得a=1.6,a=2.由于拋物線的解析式為S=(at+2)(10-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/6.png)
)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/7.png)
,經(jīng)化簡后可得出對稱軸應(yīng)該是t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/8.png)
,當(dāng)a=1.6時,對稱軸t=5.625顯然大于5,與給出的拋物線的圖形不相符,因此a=2是本題的唯一的解.也就是說P的速度是2單位/秒.
(3)根據(jù)(2)的求解過程即可得出S的解析式.然后根據(jù)函數(shù)的解析式來得出函數(shù)的最大值及此時對應(yīng)的t的取值,然后根據(jù)P,Q的速度和t的取值,可求出P點的坐標.
(4)本題其實主要是看P在B點和C點時∠OPQ的度數(shù)范圍,當(dāng)∠OBQ的度數(shù)大于90°,∠OCQ的度數(shù)小于90°時,那么在AB,BC上分別有一個符合要求的點P,如果∠OBQ的度數(shù)小于90°時那么就沒有符合要求的點,如果∠OBQ=90°,那么符合要求的點只有一個.當(dāng)P,B重合時,作∠OPM=90°交y軸于點M,作PH⊥y軸于點H,然后比較OM和OQ的長即可得出∠OPQ的大致范圍,根據(jù)相似三角形OPH和OPM不難得出OM的長,然后比較OM,OQ的大小,如果OQ>OM則說明∠OPQ>90°,反之則小于90°,用同樣的方法可得出當(dāng)P與C重合時∠OPQ的大致取值范圍,然后根據(jù)上面的分析即可判定出有幾個符合要求的點.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/images9.png)
解:(1)∵頂點B的坐標為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/9.png)
,AB=10,
∴sin∠BAO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/11.png)
,
∴∠BAO=60度.
(2)點P的運動速度為2個單位/秒.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/images13.png)
(3)過P作PM⊥x軸,
∵點P的運動速度為2個單位/秒.
∴t秒鐘走的路程為2t,即AP=2t,
又∵∠APM=30°,
∴AM=t,又OA=10,
∴OM=(10-t),即為三角形OPQ中OQ邊上的高,
而DQ=2t,OD=2,可得OQ=2t+2,
∴P(10-t,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/12.png)
t)(0≤t≤5),
∵S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/13.png)
OQ•OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/14.png)
(2t+2)(10-t),
=-(t-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/15.png)
)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/16.png)
.
∴當(dāng)t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/17.png)
時,S有最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/18.png)
,此時P(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/19.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/20.png)
).
(4)當(dāng)點P沿這兩邊運動時,∠OPQ=90°的點P有2個.
①當(dāng)點P與點A重合時,∠OPQ<90°,
當(dāng)點P運動到與點B重合時,OQ的長是12單位長度,
作∠OPM=90°交y軸于點M,作PH⊥y軸于點H,
由△OPH∽△OPM得:OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/21.png)
=11.5,
所以O(shè)Q>OM,從而∠OPQ>90度.
所以當(dāng)點P在AB邊上運動時,∠OPQ=90°的點P有1個.
②同理當(dāng)點P在BC邊上運動時,可算得OQ=12+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/22.png)
=17.8,
而構(gòu)成直角時交y軸于(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/23.png)
),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190919033718022/SYS201311011909190337180023_DA/24.png)
=20.2>17.8,
所以∠OCQ<90°,從而∠OPQ=90°的點P也有1個.
所以當(dāng)點P沿這兩邊運動時,∠OPQ=90°的點P有2個.
點評:本題結(jié)合三角形的相關(guān)知識考查二次函數(shù)的綜合應(yīng)用,要特別注意(2)中舍去速度為1.6的原因.