(i) To find the deceleration, we first calculate the frictional force using the formula:

\(F = ext{coefficient of friction} imes ext{normal force} = 0.2 imes 6g imes ext{cos}(8^ ext{o})\)

Using Newton's second law, the net force along the plane is:

\(6g imes ext{sin}(8^ ext{o}) - F = 6a\)

Solving for acceleration \(a\), we find:

\(a = rac{6g imes ext{sin}(8^ ext{o}) - F}{6}\)

Substituting the values, we get:

\(a = 0.589 ext{ m s}^{-2}\)

(ii) To find the distance traveled, we use the equation of motion:

\(v^2 = u^2 + 2as\)

Given that the final velocity \(v = 0\), initial velocity \(u = 3 ext{ m s}^{-1}\), and \(a = -0.589 ext{ m s}^{-2}\), we solve for \(s\):

\(0 = 3^2 + 2(-0.589)s\)

Solving for \(s\), we find:

\(s = 7.64 ext{ m}\)