If \(t\) is inversely proportional to \(v^2\), and \(t=11\) when \(v=2\):

Find \(t\) when \(v=4\) \begin{align}t & \propto \frac{1}{v^2}\\ So \quad t & =k (\frac{1}{v^2}) \\ By \; substitution \quad 11 & =k(\frac{1}{2^2})\\ k & =44\\ \therefore \quad t & = 44(\frac{1}{v^2})\\ And \; when \; v & =4\\ t & = 44(\frac{1}{4^2})\\ \therefore \quad t & =\frac{11}{4} \end{align}

Find \(v\) when \(t=9\) \begin{align}t & \propto \frac{1}{v^2}\\ So \quad t & =k (\frac{1}{v^2}) \\ By \; substitution \quad 11 & =k(\frac{1}{2^2})\\ k & =44\\ \therefore \quad t & = 44(\frac{1}{v^2})\\ And \; when \; t=9\\ 9 & = 44(\frac{1}{v^2})\\ v^2 & =\frac{44}{9}\\ \therefore \quad v & = \pm \sqrt{\frac{44}{9}} \end{align}