The rider experienced a centripetal acceleration of 10 g's. Given that we have gravity approximated as 9.8 m/s², then 10G = 10(9.8) = 98 m/s².
Recall that centripetal acceleration can be expressed in terms of distance and v. We have
[tex] a_{centripetal} = \frac{v^{2}}{r} [/tex]
or re-arranging the equation,
[tex]v = \sqrt{(a_{centripetal})(r)} \\ v = \sqrt{98(12)}\\ v = 34.3 m/s [/tex].
Answer: 34.3 m/s
