Đáp án:

 ô tô 2 đến trước

Giải thích các bước giải:

\(\begin{array}{l}
{t_1} = \frac{{{s_1}}}{{{v_1}}} = \frac{s}{{2{v_1}}}\\
{t_2} = \frac{{{s_2}}}{{{v_2}}} = \frac{s}{{2{v_2}}}\\
{v_{tb1}} = \frac{s}{{{t_1} + {t_2}}} = \frac{s}{{\frac{s}{{2{v_1}}} + \frac{s}{{2{v_2}}}}} = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}\\
{s_{1'}} = {v_1}{t_1}' = \frac{{{v_1}t}}{2}\\
{s_2}' = {v_2}{t_2}' = \frac{{{v_2}t}}{2}\\
{v_{tb2}} = \frac{s}{t} = \frac{{\frac{{{v_1}t}}{2} + \frac{{{v_2}t}}{2}}}{t} = \frac{{{v_1} + {v_2}}}{2}\\
{v_{tb1}} - {v_{tb2}} = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}} - \frac{{{v_1} + {v_2}}}{2} = \frac{{4{v_1}{v_2} - {{({v_1} + {v_2})}^2}}}{{2({v_1} + {v_2})}} = \frac{{ - {{({v_1} - {v_2})}^2}}}{{2({v_1} + {v_2})}}\\
 - {({v_1} - {v_2})^2} < 0 \Rightarrow {v_{tb1}} < {v_{tb2}}
\end{array}\)

suy ra ô tô 2 đến trước