a, $x_1=60t(km)\\ x_2=40+40t(km)$
b, Để hai xe gặp nhau
$60t=40+40t\\\to t=2\\ $
Thay t=2 vào $x_1\to x_1=x_2=120(km)$
c, Ta có $d=|x_1-X_2|=|60t-40t-40|=|20-40t|=120\\ \to$\(\left[ \begin{array}{l}20-40t=120\\40t-20=120\end{array} \right.\) $\to$\(\left[ \begin{array}{l}t=-2,5(L)\\t=3,5(TM)\end{array} \right.\)