Giải:
$x=2\sin (\pi t-\dfrac{\pi }{4})$
\Rightarrow $v=x'=2\pi \cos (\pi t-\dfrac{\pi }{4})$
$\left\{\begin{matrix}
x=-\sqrt{2} & \\v> 0
&
\end{matrix}\right.$
\Rightarrow $\left\{\begin{matrix}
\sin (\pi t-\dfrac{\pi }{4})=-\dfrac{1}{\sqrt{2}} & \\2\pi \cos (\pi t-\dfrac{\pi }{4})> 0
&
\end{matrix}\right.$
\Rightarrow $\left\{\begin{matrix}
\pi t-\dfrac{\pi }{4}=-\dfrac{\pi }{4}+k2\pi \\ or \ \pi t-\dfrac{\pi }{4}=\dfrac{5\pi }{4}+k2\pi & \\2\pi \cos (\pi t-\dfrac{\pi }{4})> 0
&
\end{matrix}\right.$
\Rightarrow $\pi t-\dfrac{\pi }{4}=-\dfrac{\pi }{4}+k2\pi $
\Rightarrow $t=2k$