Step 1 :Given the acceleration function \(a(t) = -6t + 2\).
Step 2 :The velocity function \(v(t)\) is the integral of the acceleration function. Integrate \(a(t)\) to get \(v(t) = -3t^2 + 2t + C_1\).
Step 3 :Given the initial condition \(v(0) = 2\), substitute into the velocity function to solve for \(C_1\). This gives \(C_1 = 2\).
Step 4 :So the velocity function is \(v(t) = -3t^2 + 2t + 2\).
Step 5 :The position function \(s(t)\) is the integral of the velocity function. Integrate \(v(t)\) to get \(s(t) = -t^3 + t^2 + 2t + C_2\).
Step 6 :Given the initial condition \(s(0) = 9\), substitute into the position function to solve for \(C_2\). This gives \(C_2 = 9\).
Step 7 :So the position function is \(s(t) = -t^3 + t^2 + 2t + 9\).
Step 8 :\(\boxed{s(t) = -t^3 + t^2 + 2t + 9}\) is the final answer.