Given:

$S=(0.5t^3+ 3t)m \\ a=10 m/s @ t= 2 sec$

Solution:

$S=0.5 t^3+ 3t \\ V= \dfrac {d_s}{dt}=0.5 ×3t^2+ 3 \\ =1.5 t^2+3 \\ a_t = \dfrac {dv}{dt}=1.5 ×2t=3t \\ At\space\space t = 2 sec \\ a_t=10 m/s^2 \\ a=\sqrt{a_t^2+ a_n^2 } \\ a_n = \sqrt{a^2- a_t^2 } =\sqrt{10^2-6^2} =64 \\ a_n=8 m/s^2 \\ V=1.5 t^2+3 \\ V=9 m/s^2\space at\space t = 2sec \\ ρ= \dfrac {V^2}{a_n} = \dfrac {9^2}8=10.125 m$

Radius of curvature = 10.125m