I got 12 :S
\frac{vdv}{dx}=-0.05v^3\newline \frac{-20dv}{v^2}=dx\newline x = \frac{20}{v}+C\newline v=10, x = 0, C=-2 \frac{20}{v}-2 = x\newline \frac{dx}{dt}=\frac{20}{x+2}\newline \frac{20dt}{dx}=x+2\newline 20t=0.5x^2+2x+C\newline x=0,t=0,C=0\newline 20t=0.5(20)^2+40\newline t = 12