(i) Using the equation of motion:

\(s = ut + \frac{1}{2}at^2\)

For AB:

\(100 = 4u + 8a\)

For BC:

\(148 = 4v + 8a\)

Using the equation \(v = u + at\) for AB to BC:

\(v = u + 4a\)

Substitute \(v = u + 4a\) into the equation for BC:

\(148 = 4(u + 4a) + 8a\)

Simplifying gives:

\(148 = 4u + 24a\)

Now solve the system of equations:

\(100 = 4u + 8a\)

\(148 = 4u + 24a\)

Subtract the first from the second:

\(48 = 16a\)

Thus, \(a = 3\) m s^-2.

Substitute \(a = 3\) into \(100 = 4u + 8a\):

\(100 = 4u + 24\)

\(76 = 4u\)

\(u = 19\) m s^-1.

(ii) Using the equation of motion:

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

For AD:

\(61^2 = 19^2 + 2 \times 3 \times s\)

\(3721 = 361 + 6s\)

\(3360 = 6s\)

\(s = 560\)

Distance CD is:

\(CD = 560 - 248 = 312\) m.