Ta có: \(\left\{{}\begin{matrix}t_1=\dfrac{s}{30}\left(h\right)\\t_2=\dfrac{\dfrac{1}{3}s}{30}+\dfrac{\dfrac{2}{3}s}{40}\left(h\right)\end{matrix}\right.\)
Theo đề bài: \(t_1-t_2=\dfrac{1}{12}\)
\(\Leftrightarrow\dfrac{s}{30}-\left(\dfrac{1}{90}s+\dfrac{1}{60}\right)s=\dfrac{1}{12}\)
\(\Leftrightarrow s=15km\)
\(\Rightarrow\left\{{}\begin{matrix}t_1=\dfrac{15}{30}=0,5h=30p\\t_2=\dfrac{\dfrac{1}{3}\cdot15}{30}+\dfrac{\dfrac{2}{3}\cdot15}{40}=\dfrac{5}{12}h=25p\end{matrix}\right.\)