Đáp án:
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
{s_1} = \dfrac{1}{2}a{t_1}^2 = \dfrac{1}{2}a{t^2}\\
{s_1} + {s_2} = \dfrac{1}{2}a{\left( {{t_1} + {t_2}} \right)^2} = \dfrac{1}{2}a.4{t^2}
\end{array} \right.\\
\Leftrightarrow \dfrac{{{s_1} + {s_2}}}{{{s_1}}} = 4 \Leftrightarrow \dfrac{{{s_1}}}{{{s_2}}} = \dfrac{1}{3}\left( 1 \right)
\end{array}$
Lại có:
$\begin{array}{l}
\left\{ \begin{array}{l}
{s_1} = \dfrac{1}{2}a{t_1}^2 = \dfrac{1}{2}a{t^2}\\
{s_1} + {s_2} + {s_3} = \dfrac{1}{2}a{\left( {{t_1} + {t_2} + {t_3}} \right)^2} = \dfrac{1}{2}a.9{t^2}
\end{array} \right.\\
\Leftrightarrow \dfrac{{{s_1} + {s_2} + {s_3}}}{{{s_1}}} = 9 \Leftrightarrow \dfrac{{{s_1}}}{{{s_1}}} + \dfrac{{{s_2}}}{{{s_1}}} + \dfrac{{{s_3}}}{{{s_1}}} = 9\\
\Leftrightarrow 1 + 3 + \dfrac{{{s_3}}}{{{s_1}}} = 9 \Rightarrow \dfrac{{{s_3}}}{{{s_1}}} = 5\left( 2 \right)
\end{array}$
Vậy: $\left( 1 \right),\left( 2 \right) \Rightarrow {s_1}:{s_2}:{s_3} = 1:3:5$
