Answer
$r(v)=\left(\frac{1}{3}+\pi\right) v^2+\frac{1}{5} v+\frac{5 \sqrt{2}-9}{15}$
Work Step by Step
We have
$$
\begin{aligned}
r(v) & =\frac{v^2+\sqrt{2}}{3}+\frac{v-3}{5}+\pi v^2 \\
& =\frac{1}{3} v^2+\frac{\sqrt{2}}{3}+\frac{v}{5}-\frac{3}{5}+\pi v^2 \\
& =\frac{1}{3} v^2+\pi v^2+\frac{1}{5} v+\frac{\sqrt{2}}{3}-\frac{3}{5} \\
& =\left(\frac{1}{3}+\pi\right) v^2+\frac{1}{5} v+\frac{5 \sqrt{2}-9}{15}
\end{aligned}
$$
so $a=1 / 3+\pi, b=1 / 5, c=(5 \sqrt{2}-9) / 15$
