(i) The acceleration during the first 2 seconds is the gradient of the velocity-time graph from \(t = 0\) to \(t = 2\). The change in velocity is from \(0\) to \(-2\) m s^-1, so the acceleration \(a\) is given by:

\(a = \frac{-2 - 0}{2 - 0} = -1 \text{ m s}^{-2}\).

(ii) To find \(V\), use the gradient of the line between \(t = 4\) and \(t = 10\). The change in velocity is from \(-2\) to \(V\), and the change in time is \(6\) seconds:

\(\frac{V + 2}{6} = 1\).

Solving for \(V\), we get:

\(V + 2 = 6\)

\(V = 4\).

(iii) The distance \(AB\) is three times the distance traveled in the first 6 seconds. The area under the graph from \(t = 0\) to \(t = 6\) is:

\(\frac{1}{2} \times (6 + 2) \times 2 = 8 \text{ m}\).

Thus, \(AB = 3 \times 8 = 24 \text{ m}\).

For \(t = T\), the area under the graph from \(t = 6\) to \(t = T\) must also be \(16 \text{ m}\) (since \(AB = 24 \text{ m}\) and \(8 \text{ m}\) is already covered):

\(\frac{1}{2} \times (T - 6) \times 4 = 16\).

Solving for \(T\), we get:

\(2(T - 6) = 16\)

\(T - 6 = 8\)

\(T = 18\).