(2) According to the conditions that congruent triangles should meet, explore the relationship between edges, and then according to the formula of distance = speed × time, first find out the moving time of point P, and then find out the moving speed of point Q;
(2) According to the meaning of the question and graphic analysis, it is found that point Q is faster than point P, so if we want to meet for the first time, we should take two sides of an equilateral triangle longer than point P. 。
Solution: solution: (1)①∫t = 1 sec,
∴ 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，
∴△BPD≌△CPQ.
②∵vP≠vQ,∴BP≠CQ，
And ∵△BPD?△CPQ, ∠B=∠C, BP=PC=4, CQ=BD=5,
∴ point p, point q moved by t=BP/3=4/3 seconds,
∴VQ = CQ/t = 5/(4/3)= 15/4cm/s；
(2) suppose that point p and point q meet for the first time after x seconds,
From the meaning of the title, it is 15/4x = 3x+2x 10.
The solution is x=80/3 seconds.
∴ The point moved by 80/3×3=80 cm.
∵80═56+24=2×28+24,
Point p and point q meet on the side of AB,
After 80/3 seconds, point P and point Q meet for the first time on the AB side.