3, we can get AB = 2OB. ..

In Rt△AOB, OB= 12 and ab = 24 are obtained by Pythagorean theorem.

∴B(0, 12)，

∫OA = 12

3,

∴A ( 12

3,0).

The analytical formula of straight line AB can be set as: y=ax+b,

∴b= 12 12

3a+b=0？ ,

∴a=-

33b= 12？ ,

∴ The analytical formula of available straight line AB is: y=-

33x+ 12。

(2) Connect the CD, if F is the FM⊥x axis of point M, then CB = CD.

∵∠OBA=90 -∠A=60，

△ CBD is an equilateral triangle.

∴BD=CB= 12OB=6，

∠BCD=60，∠OCD= 120。

∫ob is the diameter, OA⊥OB,

∴OA Cut C to O

⊙ dechuts ⊙ c to d,

Coe = CDE = 90, Organization of Eastern Caribbean States =.

∴∠OED=360 -∠COE-∠CDE-∠OCD=60。

∴∠OEC=∠DEC=30。

∴CE= 12,CO=6.

∴ in Rt△COE, Pythagorean Theorem OE=CE2-CO2=6.

3.

∵BG⊥EC in F,

∴∠GFE=90。

∠∠GBO+∠BGO =∠OEC+∠BGO，

∴∠GBO=∠OEC=30。

Therefore, FC= 12BC=3, EF=FC+CE= 15,

FM= 12EF= 152，ME=3FM= 15

32.

∴MO= 15

32-6

3=

three

32.

∴F(-

three

32, 152).

(3) Set the moving speed of Q point as VCM/s. 。

(i) When point P moves to the midpoint of AB, point Q moves to the midpoint of AO, PQ∑BC, PQ=BC, the quadrilateral CBPQ is a parallelogram, and point Q coincides with point E. 。

Available AP= 12, t=

AP4=3。

∴v=

AEt=

six

33=2

3 (cm/s).

(ii) When point P moves to BG midpoint and point Q moves to OG midpoint,

PQ∑BC, PQ=BC, quadrilateral CBPQ is a parallelogram.

Available OG=4

3，BG=8

3. Therefore PB=4.

3，OQ=2

3.

∴t=

AB+BP4=

24+4

34=6+

3.

∴v=

AQt=

12

3+2

36+

3=

28

3- 14 1 1 (cm/s).

The speed of point Q is 2.

3 cm/s or 28

3-1411cm/s comments: This topic mainly examines the comprehensive application of linear functions, the properties of parallelograms and the application of Pythagorean theorem, and classifies them according to the position of known point P motion.