(1)AB = 11,0 ≤ t ≤ 11t秒时,P(-8 + t,0),BQ = tOB = 3,OC = 4,BC = 5cos∠ABC = OB/BC = 3/5; sin∠ABC = OC/BC = 4/5Q的横坐标 = OB - BQcos∠ABC = 3 - 3t/5Q的纵坐标 = BQsin∠ABC = 4t/5当PQ⊥x轴时,P,Q的横...第3题的3是如何求出的，第4题能用初中知识吗AQ² = (3 - 3t/5 + 8)² + (4t/5 - 0)² = (11 - 3t/5)² + (4t/5)²PQ² = (-8 + t - 3 + 3t/5)² + (4t/5)² = (8t/5 - 11)² + (4t/5)(iii)Q为顶点, AQ² = PQ²(11 - 3t/5)² + (4t/5)² = (8t/5 - 11)² + (4t/5)²(11 - 3t/5)²= (8t/5 - 11)²11 - 3t5 = 8t/5 - 11, 11t/5 = 22, t = 10或11 - 3t/5 = 11 - 8t/5, t = 0, 舍去