解:设t秒钟后,p、Q间的距离等于4√2此时 AP=t, BQ=2t, PB=AB-AP=6-tPQ²=BP²+BQ²(4√2)²=(6-t)²+(2t)²5t²-12t+4=0(5t-2)(t-2)=0t=2 (2t=4cm >BC 舍去) t=2/5=0.4答:0.4秒钟后,p、Q间的距离等于4倍根号2