Then d= 1 1-2t.

(2) From the meaning of the question, the tangent point of the two circles is on the straight line MN, and the arc distance of the original state of the two circles is 9.

Let t(s) be the tangent of two circles.

The distance that the outer arc of circle A moves is d 1=2t.

The moving distance of the outer arc of circle B is D2 =1+t-1= t.

The expression of the distance between two arcs is d = 9-(d 1+D2) = 9-3t.

When d = 0, the two circles are tangent, that is, 9-3t = 0.

T = 3 (seconds)