You can figure it out yourself. When t=4, the length of equilateral triangle is 8, the height is similar, and the length covering AD is 2. From the second question, we can know that when t=5, the coverage length of AD is also 2, so we can know that during the period of 4≤t≤5, the coverage length of AD is always 2 (because after t > 5, the equilateral triangle and AD.

This is my thinking process in the examination room. I may not remember clearly.