$ 解：當(dāng)0＜t≤\frac{2}{3}時(shí)$
$S=\frac{16}{9}-\frac{1}{2}×t×2t=-t^2+\frac{16}{9}$
$當(dāng)\frac{2}{3}＜t≤\frac{4}{3}時(shí)$
$S=\frac{1}{2}[（\frac{4}{3}-t）+（2-t）]×\frac{4}{3}=-\frac{4}{3}t+\frac{20}{9}$
$當(dāng)\frac{4}{3}＜t≤2時(shí)$
$S=\frac{1}{2}×（2-t）×（4-2t）=t^2-4t+4$
