\(4m/s=\dfrac{72}{5}\left(\dfrac{km}{h}\right),1h20ph=\dfrac{4}{3}h\)
\(\left\{{}\begin{matrix}S_1=v_1.t=\dfrac{4}{3}.\dfrac{72}{5}=\dfrac{96}{5}\left(km\right)\\S_2=v_2.t=\dfrac{4}{3}.36=48\left(km\right)\end{matrix}\right.\)
\(S_{AB}=S_1+S_2=48+\dfrac{96}{5}=67,2\left(km\right)\)