22.如圖,已知△ABC是邊長(zhǎng)為12cm的等邊三角形,動(dòng)點(diǎn)P,Q同時(shí)從AB兩點(diǎn)出發(fā),分別沿AB、BC勻速運(yùn)動(dòng),其中點(diǎn)P運(yùn)動(dòng)的速度是2cm/s,點(diǎn)Q運(yùn)動(dòng)的速度是4cm/s,當(dāng)點(diǎn)Q到達(dá)點(diǎn)C時(shí),P、Q兩點(diǎn)都停止運(yùn)動(dòng),設(shè)運(yùn)動(dòng)時(shí)間為t(s),解答下列問(wèn)題:
(1)當(dāng)t=2時(shí),判斷△BPQ的形狀,并說(shuō)明理由;
(2)設(shè)△BPQ的面積為S(cm
2),求S與t的函數(shù)關(guān)系式;
(3)作QR∥BA交AC于點(diǎn)R,連接PR,當(dāng)t為何值時(shí),△APR∽△PRQ.