(3) When point N moves on DA, the speed at which point P moves to B is equal to the speed at which point N moves to A,

Because qp/BP = tan 30 = √ 3/3,

So the speed of QP reduction is √3T/3, that is, QP=√3-√3T/3,

So s = (1/2) * BM * qp = (1/2) * t * (√ 3-√ 3t/3) = √ 3t/2-√ 3t2/6, (0

When point N moves on AB, the speed of point P moving to B is equal to COS60 * 1 = 1/2.

Because qp/BP = tan 30 = √ 3/3,

So the speed of QP reduction is (√ 3/3) * (1/2) * (t-2) = √ 3 (t-2)/6.

When n is at point A, BP= 1, so QP=√3/3.

Therefore, the moving length of QP is =√3/3-√3(T-2)/6=√3(4-T)/6.

So s = (1/2) * BM * qp = (1/2) * t * √ 3 (4-t)/6 = √ 3t/3-√ 3t2/12, (2

So s = √ 3t2/2-√ 3t2/6, (0

=√3T/3-√3T^2/ 12,(2≤T≤4)

(4).T=3/2, or T= 12/7.