Đáp án:
\(5s\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{4}{5}{v_0} = {v_0} + a.t \Rightarrow - \dfrac{{{v_0}}}{5} = a\\
{\left( {\dfrac{4}{5}{v_0}} \right)^2} - v_0^2 = 2a.AB\\
\Rightarrow - \dfrac{9}{{25}}v_0^2 = - 2.\dfrac{{{v_0}}}{5}.AB\\
\Rightarrow AB = \dfrac{9}{{10}}{v_0}\\
v_C^2 - v_0^2 = 2a.AC\\
\Rightarrow v_C^2 - v_0^2 = - 2.\dfrac{{{v_0}}}{5}.\dfrac{{25}}{9}.\dfrac{9}{{10}}{v_0}\\
\Rightarrow v_C^2 = 0 \Rightarrow {v_C} = 0\\
{t_C} = \dfrac{{{v_C} - {v_0}}}{a} = \dfrac{{ - {v_0}}}{{ - \dfrac{{{v_0}}}{5}}} = 5s
\end{array}\)
