## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Let the centripetal acceleration $a_c$ be $10~g$. $a_c = \frac{v^2}{r}$ $10~g = \frac{v^2}{r}$ $v^2 = 10~g~r$ $v = \sqrt{10~g~r}$ $v = \sqrt{(10)(9.80~m/s^2)(12~m)}$ $v = 34.3~m/s$ The rider is moving with a speed of 34.3 m/s.