∴ BP = CQ = 3× 1 = 3cm，

∫AB = 10cm, and point D is the midpoint of AB.

∴BD=5 cm

∫PC = BC∫BP，BC=8 cm，

∴ PC = 8 3 = 5 cm,

∴PC=BD.

AB = AC，

∴∠B=∠C，

At △BPD and △CQP,

∴△BPD≌△CQP.(SAS)

②∵vP≠vQ,∴BP≠CQ，

And ∵△BPD?△CPQ, ∠B=∠C, BP=PC=4cm, CQ=BD=5cm,

Point p, point q moves in seconds,

∴ cm/sec;

(2) suppose that point p and point q meet for the first time after x seconds,

From the meaning of the question, x = 3x+2x 10,

Solve.

∴ Point P*** has moved by ×3=80 cm.

∴80=56+24=2×28+24,

Point p and point q meet on the side of AB,

A second later, point P and point Q meet at the edge of AB for the first time. Hope to adopt.