1.2.21 | revision #13617 | Savchenko Solutions

Statement

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?

Solution

$l=\nu_{1}t_{1} +\frac{at_{1}^{2}}{2}$

$l=(\nu_{1}+at_{1})t_{2} +\frac{at_{2}^{2}}{2}$

$(\nu_{1}+at_{1})t_{2} +\frac{at_{2}^{2}}{2} = \nu_{1}t_{1} +\frac{at_{1}^{2}}{2}$

Given that $\nu_{1} = a{d}t$ :

$a{d}t(t_{1}-t_{2}) =\frac{a(t_{2}^{2}+2t_{1}t_{2}-t_{1}^{2})}{2}$

${d}t=\fbox{$\frac{t_{2}^{2}+2t_{1}t_{2}-t_{1}^{2}}{2(t_{1}-t_{2})}$}$

Answer

$\frac{t_{2}^{2}+2t_{1}t_{2}-t_{1}^{2}}{2(t_{1}-t_{2})}$