1.如圖,在四邊形ABCD中,AB=BC=CD=BD=5厘米,動(dòng)點(diǎn)P從點(diǎn)A出發(fā)沿AB→BD運(yùn)動(dòng)到點(diǎn)D,速度為2厘米/秒,動(dòng)點(diǎn)Q從點(diǎn)D出發(fā)沿DC→CB→BA運(yùn)動(dòng)到點(diǎn)A,速度為2.8厘米/秒.設(shè)運(yùn)動(dòng)時(shí)間為t,當(dāng)t=5秒時(shí),P,Q兩點(diǎn)相距3厘米.
(1)當(dāng)t=5秒時(shí),P,Q兩點(diǎn)的運(yùn)動(dòng)路程分別是多少?
(2)當(dāng)t=5秒時(shí),試判斷△APQ的形狀,并說(shuō)明理由.