Answer
.1 percent
Work Step by Step
Plugging in $v=\omega A$ and $\omega=2\pi f$ into the equation $E=\frac{1}{2}mv^2$, we find:
$E = 2\pi^2mf^2A^2 \\ E= 2\pi^2(1400)(.67)^2(.18)^2 =401.93 \ J$
We divide this by the kinetic energy of the car to find:
$= \frac{401.6\ J}{\frac{1}{2}mv^2}=\frac{401.6\ J}{\frac{1}{2}(1400)(20)^2}=.001 \approx\fbox{.1 percent}$
