【答案】
分析:(1)已知了AB的長和B點的坐標(biāo),那么sin∠BAO=
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=
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,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
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a,也就是P到OQ的距離為10-
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a.OQ=QD+OD=5a+2.因此(5a+2)×(10-
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)×
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=30,解得a=1.6,a=2.由于拋物線的解析式為S=(at+2)(10-
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)×
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,經(jīng)化簡后可得出對稱軸應(yīng)該是t=
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,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(biāo).
(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:
解:(1)∵頂點B的坐標(biāo)為
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,AB=10,
∴sin∠BAO=
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=
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,
∴∠BAO=60度.
(2)點P的運動速度為2個單位/秒.
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(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,
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t)(0≤t≤5),
∵S=
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OQ•OM=
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(2t+2)(10-t),
=-(t-
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)
2+
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.
∴當(dāng)t=
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時,S有最大值為
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,此時P(
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,
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).
(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=
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=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+
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=17.8,
而構(gòu)成直角時交y軸于(0,
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),
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=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.