(i) The acceleration between \(t = 3.5\) and \(t = 6\) is given by the change in velocity divided by the time interval. The velocity changes from \(10 \text{ m s}^{-1}\) to \(-15 \text{ m s}^{-1}\) over \(2.5 \text{ s}\). Thus, the acceleration is:

\(a = \frac{-15 - 10}{6 - 3.5} = \frac{-25}{2.5} = -10 \text{ m s}^{-2}\)

(ii) The acceleration between \(t = 6\) and \(t = 10\) is \(7.5 \text{ m s}^{-2}\). The initial velocity at \(t = 6\) is \(-15 \text{ m s}^{-1}\). The velocity at \(t = 10\) is \(V \text{ m s}^{-1}\). Using the formula:

\(V = -15 + 7.5 \times 4\)

\(V = 15 \text{ m s}^{-1}\)

(iii) The particle comes to rest at \(t = T\). The total distance travelled is \(100 \text{ m}\). The areas under the velocity-time graph represent the distance travelled. The areas are calculated as follows:

Area from \(t = 0\) to \(t = 4.5\):

\(\frac{1}{2} \times (4.5 + 2) \times 10\)

Area from \(t = 4.5\) to \(t = 8\):

\(\frac{1}{2} \times (8 - 4.5) \times 15\)

Area from \(t = 8\) to \(t = T\):

\(\frac{1}{2} \times (T - 8) \times 15\)

Setting the total distance to \(100 \text{ m}\):

\(\frac{1}{2} \times (4.5 + 2) \times 10 + \frac{1}{2} \times (8 - 4.5) \times 15 + \frac{1}{2} \times (T - 8) \times 15 = 100\)

Solving for \(T\):

\(T = 13.5 \text{ s}\)