$1.2.21^*.$ The scheduled departure time of the train is $12.00$. It's $12.00$ on your watch, but the penultimate car that moves past you during time $t_1$ is already starting to pass by you. The last car passes you during $t_2$. The train left on time and is moving equidistant. How far behind is your watch?
$$l=\nu _{1}t_{1} +\frac{at_{1}^{2}}{2}$$ $$l=(\nu _{1}+at_{1})t_{2} +\frac{at_{1}^{2}}{2}$$ $$(\nu _{1}+at_{1})t_{2} +\frac{at_{1}^{2}}{2} = \nu _{1}t_{1} +\frac{at_{1}^{2}}{2}$$ Given that $\nu _{1} = a{d}t$: $$a{d}t(t_{1}-t_{2}) =\frac{a(t_{2}-2t_{1}t_{2}-t_{1}^{2})}{2}$$ $${d}t=\fbox{$\frac{t_{2}-2t_{1}t_{2}-t_{1}^{2}}{2(t_{1}-t_{2})}$}$$
$$\frac{t_{2}-2t_{1}t_{2}-t_{1}^{2}}{2(t_{1}-t_{2})}$$