∴ As long as QC=PD, the quadrilateral PQCD is a parallelogram.

At this time, 3t = 24,

T=6。

That is, when t=6 seconds, the quadrilateral PQCD is a parallelogram.

Similarly, as long as PQ=CD, PD≠QC,

The quadrilateral PQCD is an isosceles trapezoid.

The vertical lines passing through P and D respectively are BC and intersect E and F (as shown in Figure 2-2), so we can know from the properties of isosceles trapezoid:

EF=PD，QE=FC=2 .

∴2=[3t(24-t)]

The solution is t=7.

∴t=7 seconds, the quadrilateral PQCD is an isosceles trapezoid.

(2) When the movement lasts for t seconds, the straight lines PQ and ⊙O are tangent to point G (as shown in Figure 2-3), and P is taken as PH⊥BC and the vertical foot is H.

Then PH=AB, BH=AP,

That is, PH=8, HQ=26-3t-t=26-4t.

According to the tangent length theorem, PQ=AP+BQ=t+26-3t=26-2t.

According to Pythagorean theorem, PQ2=PH2+HQ2,

That is, (26-2t)2=82+(26-4t)2.

Simplified sorting, 3t2-26t+ 16=0.

Solution, t 1= t2=8.

That is, when t= seconds or t=8 seconds, the straight line PQ is tangent to ⊙ O.

When ∵t=0 (seconds), PQ intersects ⊙ o; When t==8 (seconds), point Q moves to point B, and point P does not move to point D, but it also stops moving. At this time, PQ also intersects with ⊙ O.

∴ When t= or t=8, the straight line PQ is tangent to⊙ O;

When 0 ≤ t

Or 8 < t ≤ 8, and straight lines PQ intersect ⊙ O;

What time? When < t < 8, the straight line PQ and ⊙O are separated.