Đáp án:
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
{A^2} = \dfrac{{{v^2}}}{{{\omega ^2}}} + \dfrac{{{a^2}}}{{{\omega ^4}}} \Leftrightarrow {\omega ^2}{A^2} = {v^2} + \dfrac{{{a^2}}}{{{\omega ^2}}}\\
{v_{\max }}^2 = {\omega ^2}{A^2}
\end{array} \right.\\
\Rightarrow {v_{\max }}^2 = {v^2} + {\left( {\dfrac{a}{\omega }} \right)^2}
\end{array}$