Đáp án:
${v_{tb}} = 30km/h$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{t_1} = \dfrac{{{s_1}}}{{{v_1}}} = \dfrac{s}{{3{v_1}}} = \dfrac{s}{{3.60}} = \dfrac{s}{{180}}\\
{t_2} = \dfrac{{{s_2}}}{{{v_2}}} = \dfrac{{{s_2}}}{{{v_{tb}}}}\\
{t_3} = \dfrac{t}{3} = \dfrac{s}{{3{v_{tb}}}}
\end{array}$
Lại có:
$\begin{array}{l}
{s_2} + {s_3} = \dfrac{2}{3}s \Leftrightarrow {v_2}{t_2} + {v_3}{t_3} = \dfrac{2}{3}s\\
\Leftrightarrow {v_{tb}}{t_2} + \dfrac{s}{{3{v_{tb}}}}.15 = \dfrac{2}{3}s \Leftrightarrow {v_{tb}}{t_2} + \dfrac{{5s}}{{{v_{tb}}}} = \dfrac{2}{3}s \Leftrightarrow {v_{tb}}{t_2} = \dfrac{{2s}}{3} - \dfrac{{5s}}{{{v_{tb}}}}\\
\Leftrightarrow {t_2} = \dfrac{{2s}}{{3{v_{tb}}}} - \dfrac{{5s}}{{{v_{tb}}^2}} = \dfrac{{2s{v_{tb}} - 15s}}{{3{v_{tb}}^2}} = \dfrac{{s\left( {2{v_{tb}} - 15} \right)}}{{3{v_{tb}}^2}}
\end{array}$
Vận tốc trung bình của xe trên cả quãng đường là:
$\begin{array}{l}
{v_{tb}} = \dfrac{s}{{{t_1} + {t_2} + {t_3}}} = \dfrac{s}{{\dfrac{s}{{180}} + \dfrac{{s\left( {2{v_{tb}} - 15} \right)}}{{3{v_{tb}}^2}} + \dfrac{s}{{3{v_{tb}}}}}}\\
\Leftrightarrow {v_{tb}} = \dfrac{1}{{\dfrac{1}{{180}} + \dfrac{{2{v_{tb}} - 15}}{{3{v_{tb}}^2}} + \dfrac{1}{{3{v_{tb}}}}}} \Leftrightarrow {v_{tb}} = \dfrac{1}{{\dfrac{1}{{180}} + \dfrac{{3{v_{tb}} - 15}}{{3{v_{tb}}^2}}}}\\
\Leftrightarrow \dfrac{1}{{{v_{tb}}}} - \dfrac{{3{v_{tb}} - 15}}{{3{v_{tb}}^2}} = \dfrac{1}{{180}} \Leftrightarrow \dfrac{{3{v_{tb}} - 3{v_{tb}} + 15}}{{3{v_{tb}}^2}} = \dfrac{1}{{180}}\\
\Leftrightarrow \dfrac{{15}}{{{v_{tb}}^2}} = \dfrac{1}{{60}} \Leftrightarrow {v_{tb}}^2 = 900 \Rightarrow {v_{tb}} = 30km/h
\end{array}$
