## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Let the centripetal acceleration $a_c$ be $1.5~g$; $a_c = \frac{v^2}{r}$ $1.5~g = \frac{v^2}{r}$ $r = \frac{v^2}{1.5~g}$ $r = \frac{(25~m/s)^2}{(1.5)(9.80~m/s^2)}$ $r = 42.5~m$ The radius of the turn is 42.5 meters.