Chapter 12 - Kinematics of a Particle - Section 12.7 - Curvilinear Motion: Normal and Tangential Components - Problems - Page 65: 114

$\rho=1280$ ft $a_t=8.66$ ft/s^2

The tangential acceleration is: $a_t=a\cos 30 = 10 \cos 30 = 8.66 ft/s^2$ and the normal acceleration is $a_n=a\sin 30 = 10 \sin 30 = 5.00 ft/s^2$. Applying $a_n=\frac{v^2}{\rho}$, we have $\rho=\frac{v^2}{a_n}=\frac{80^2}{5.00}=1280$ ft