解：設(shè)運動時間為?$ts，$?顯然?$0\leq t\leq 3.$?
?$(1)$?過點?$Q_{作}QE\perp AB$?于點?$E，$?過點?$A$?作?$AF\perp CD$?于點?$F.$?
∵?$AB// CD，$??$∠C = 90°，$??$AF\perp CD，$??$QE\perp AB，$?
∴易得四邊形?$AFCB$?和四邊形?$AFQE$?都為矩形?$.$?
∵?$CD = 10\ \mathrm {cm}，$??$AB = 6\ \mathrm {cm}，$?∴?$CF = 6\ \mathrm {cm}，$?則?$DF = 4\ \mathrm {cm}.$?
∵?$AD = 5\ \mathrm {cm}，$?
∴在?$Rt\triangle ADF_{中}，$?根據(jù)勾股定理?$AF=\sqrt {AD^2-DF^2}=\sqrt {5^2-4^2}=\sqrt {25 - 16}=3\ \mathrm {cm}，$?
∴?$EQ = AF = 3\ \mathrm {cm}.$?
∵?$AP = 2t\mathrm {cm}，$??$CQ = t\mathrm {cm}，$?
∴易得?$PE=(6 - 3t)\mathrm {cm}_{或}PE=(3t - 6)\mathrm {cm}.$?
?$ $?在?$Rt\triangle PEQ_{中}，$?∵?$PE^2+EQ^2=PQ^2，$?
∴?$(6 - 3t)^2+3^2=5^2，$?
?$ 36-36t + 9t^2+9 = 25，$?
?$ 9t^2-36t + 20 = 0，$?
?$ $?對于一元二次方程?$ax^2+bx + c = 0(a\neq 0)，$?這里?$a = 9，$??$b = - 36，$??$c = 20，$?
?$ $?根據(jù)求根公式?$x=\frac {-b\pm \sqrt {b^2-4ac}}{2a}，$?可得
?$t=\frac {36\pm \sqrt {(-36)^2-4×9×20}}{2×9}=\frac {36\pm \sqrt {1296 - 720}}{18}=\frac {36\pm \sqrt {576}}{18}=\frac {36\pm 24}{18}，$?
?$ $?解得?$t_{1}=\frac {36 + 24}{18}=\frac {10}{3}，$??$t_{2}=\frac {36 - 24}{18}=\frac {2}{3}，$?
?$ $?因為?$0\leq t\leq 3，$?所以?$t_{2}=\frac {2}{3}$?符合題意，?$t_{1}=\frac {10}{3}$?不合題意，舍去.
答：經(jīng)過?$\frac {2}{3}s，$?點?$P、$??$Q $?之間的距離為?$5\ \mathrm {cm}.$?
?$ (2)$?不存在?$.$?
理由：假設(shè)存在某一時刻，使得?$PD$?恰好平分?$∠APQ，$?則?$∠APD=∠DPQ.$?
∵?$AB// CD，$?∴?$∠APD=∠PDQ，$?
∴?$∠PDQ=∠DPQ，$?∴?$DQ = PQ.$?
∵?$PQ^2=[3^2+(6 - 3t)^2]\mathrm {cm}2，$??$DQ^2=(10 - t)^2\ \mathrm {cm}2，$?
∴?$3^2+(6 - 3t)^2=(10 - t)^2，$?
?$ 9 + 36-36t + 9t^2=100-20t + t^2，$?
?$ 9t^2-t^2-36t + 20t+9 + 36 - 100 = 0，$?
?$ 8t^2-16t - 55 = 0，$?
?$ $?對于一元二次方程?$ax^2+bx + c = 0(a\neq 0)，$?這里?$a = 8，$??$b = - 16，$??$c = - 55，$?
?$ $?根據(jù)求根公式?$t=\frac {-b\pm \sqrt {b^2-4ac}}{2a}=\frac {16\pm \sqrt {(-16)^2-4×8×(-55)}}{2×8}=\frac {16\pm \sqrt {256 + 1760}}{16}=\frac {16\pm \sqrt {2016}}{16}=\frac {16\pm 12\sqrt {14}}{16}=\frac {4\pm 3\sqrt {14}}{4}，$?
∵?$0\leq t\leq 3，$?上述兩解均不合題意，舍去，
∴不存在某一時刻，使得?$PD$?恰好平分?$∠APQ.$?
