(i) To find the acceleration, differentiate the velocity functions:

For \(0 \leq t \leq 5\), \(v(t) = 0.72t^2 - 0.096t^3\).

\(a_1(t) = \frac{d}{dt}(0.72t^2 - 0.096t^3) = 1.44t - 0.288t^2\).

For \(5 \leq t \leq 10\), \(v(t) = 2.4t - 0.24t^2\).

\(a_2(t) = \frac{d}{dt}(2.4t - 0.24t^2) = 2.4 - 0.48t\).

Evaluate at \(t = 5\):

\(a_1(5) = 1.44 \times 5 - 0.288 \times 25 = 0\).

\(a_2(5) = 2.4 - 0.48 \times 5 = 0\).

Since \(a_1(5) = a_2(5) = 0\), there is no instantaneous change in acceleration.

(ii) To find the distance \(AB\), integrate the velocity functions:

For \(0 \leq t \leq 5\), \(s_1 = \int_0^5 (0.72t^2 - 0.096t^3) \, dt = [0.24t^3 - 0.024t^4]_0^5\).

\(s_1 = (0.24 \times 5^3 - 0.024 \times 5^4) - (0 - 0) = 15\).

For \(5 \leq t \leq 10\), \(s_2 = \int_5^{10} (2.4t - 0.24t^2) \, dt = [1.2t^2 - 0.08t^3]_5^{10}\).

\(s_2 = (1.2 \times 10^2 - 0.08 \times 10^3) - (1.2 \times 5^2 - 0.08 \times 5^3) = 20\).

Total distance \(AB = s_1 + s_2 = 15 + 20 = 35\) m.