Ta có ${{\left( \frac{v}{{{v}_{\max }}} \right)}^{2}}+{{\left( \frac{a}{{{a}_{\max }}} \right)}^{2}}=1$ nên ${{\left( \frac{15\pi \sqrt{3}}{30\pi \sqrt{3}} \right)}^{2}}+{{\left( \frac{22,5}{{{a}_{\max }}} \right)}^{2}}=1$ $\to $ ${{a}_{\max }}=15\sqrt{3}\,m/{{s}^{2}}$.Vậy $\left\{ \begin{align}& A=\frac{v_{\max }^{2}}{{{a}_{\max }}}=\frac{{{\left( 30\pi \sqrt{3} \right)}^{2}}}{1500\sqrt{3}}=6\sqrt{3}\,cm \\& \omega =\frac{{{a}_{\max }}}{{{v}_{\max }}}=\frac{1500\sqrt{3}}{30\pi \sqrt{3}}=5\pi \,rad/s \\\end{align} \right.$, nên $18\sqrt{3}=3A=2A+A$ $\to {{t}_{\max }}=\frac{T}{2}+2.\frac{T}{6}=\frac{5T}{6}=\frac{1}{3}\,s$.