As shown in the figure, △ABC is an equilateral triangle with a side length of 3cm. Moving points P and Q start from point A and B at the same time and move in the direction of AB and BC at a uniform speed, both of which are1cm/s. When point P reaches point B, points P and Q stop moving. Let the moving time of point P be t(s) and answer the following questions:

(1) When t is what value, is △PBQ a right triangle?

(2) Let the area of quadrilateral APQC be y(cm2) and find the relationship between y and t; Is there a moment t that makes the area of quadrilateral APQC two-thirds of the area of △ABC? If it exists, find the corresponding t value; If it does not exist, explain the reasons;

(3) Let the length of PQ be x(cm), and try to determine the relationship between Y and X. 。

Solution: (1) According to the meaning of the question: AP = t cm, BQ = t cm. △ ABC，AB = BC = 3 cm，∠ B = 60，∴ BP = (3-t) cm。

△ in△ △PBQ, BP = 3-t, bq = t,?

If △PBQ is a right triangle, then ∠ bqp = 90 or ∠ bpq = 90.

When ∠ bqp = 90, bqp = 1/2bp.

That is t = 1/2 (3-t),?

T = 1 (seconds).

When ∠ bpq = 90, BP = 1/2bq.

That is 3-t = 1/2t,?

T = 2 (seconds).

Answer: When t = 1 sec or t = 2 sec, △PBQ is a right triangle .4'