Đáp án:
$\left[ \begin{array}{l}
a = - 0,9m/{s^2}\\
a = - 0,1m/{s^2}
\end{array} \right.$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{s_1} = {v_o}{t_1} + \dfrac{1}{2}a{t^2} \Leftrightarrow 60 = 20{v_o} + \dfrac{1}{2}{.20^2}a = 20{v_o} + 200a\\
\Leftrightarrow a = \dfrac{{60 - 20{v_o}}}{{200}} = \dfrac{{3 - {v_o}}}{{10}}
\end{array}$
Ta lại có:
$\begin{array}{l}
a = \dfrac{{{v^2} - {v_o}^2}}{{2{s_2}}} = \dfrac{{{0^2} - {v_o}^2}}{{2.80}} \Leftrightarrow - 160a = {v_o}^2\\
\Leftrightarrow - 160.\dfrac{{3 - {v_o}}}{{10}} = {v_o}^2 \Leftrightarrow - 48 + 16{v_o} = {v_o}^2\\
\Leftrightarrow {v_o}^2 - 16{v_o} + 48 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{v_o} = 12m/s\\
{v_o} = 4m/s
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
a = \dfrac{{3 - {v_o}}}{{10}} = \dfrac{{3 - 12}}{{10}} = - 0,9m/{s^2}\\
a = \dfrac{{3 - {v_o}}}{{10}} = \dfrac{{3 - 4}}{{10}} = - 0,1m/{s^2}
\end{array} \right.
\end{array}$
