A

Step 1: For a point in circular motion, the normal acceleration is given by \( n = \frac{v^2}{R} \) and tangential acceleration is given by \( a = \frac{dv}{dt} \).
Step 2: We're given that \( a = a \) (constant) and \( n = bt^4 \). Since normal acceleration depends on velocity and the radius of curvature, we can relate this to the tangential acceleration via the equation for total acceleration: \( A = \sqrt{a^2 + n^2} \).
Step 3: Integrate the tangential acceleration to find the velocity function. From \( a = a \), we have \( v(t) = at \) (given that it starts from rest).
Step 4: Substitute velocity into the normal acceleration and solve for \( R \): \( R = \frac{(at)^2}{bt^4} = \frac{a^2}{b} t^{-2} \).
Finally, consider the total acceleration \( A = \sqrt{a^2 + n^2} \) to express it in terms of distance \( s \) which can be obtained from \( s = \int v(t) dt = \int at dt = \frac{1}{2} a t^2 \). Thus, express \( t \) in terms of \( s \): \( t = \sqrt{\frac{2s}{a}} \). Substitute it back to get expressions for both \( R \) and \( A \) in terms of \( s \).
Therefore, the answers will vary with distance covered \( s \), and thus option A (as suggested) captures the dependency.