The point $\left(0\,-2\right)$ is on the branch of the ellipse defined by $y\left(x\right)\= -2\sqrt{1-{\left(x\/3\right)}^2}$, which has curvature

$\mathrm{\κ}\=\frac{\|y″\|}{{\left(1\+{\left(y\prime \right)}^2\right)}^{3\/2}}$ = $\frac{162}{{\left(81-5 x^2\right)}^{3\/2}}$ 

At $x\=0$, $\mathbf{T}\left(0\right)\=\mathbf{i}$, $\mathbf{N}\left(0\right)\=\mathbf{j}$, $\mathrm{\κ}\left(0\right)\=162\/{81}^{3\/2}\=2\/9$, and the equation $\mathbf{F}\=m \mathbf{a}$ becomes

$\left[\begin{array}{c}3\\ 5\end{array}\right]$ = $3\(\stackrel{\.}{v} \left[\begin{array}{c}1\\ 0\end{array}\right]\+\frac{2}{9} v^2 \left[\begin{array}{c}0\\ 1\end{array}\right]\)$ 

from which the equation $5\=2 v^2\/3$ can be extracted. Since the speed must be positive, $v\=\sqrt{15\/2}$.