(i) To find the acceleration, differentiate the velocity function \(v(t) = 0.00004t^3 - 0.006t^2 + 0.288t\) with respect to \(t\):

\(a(t) = \frac{dv}{dt} = 0.00012t^2 - 0.012t + 0.288\).

Set \(a(t) = 0\) to find when the acceleration is zero:

\(0.00012t^2 - 0.012t + 0.288 = 0\).

Factor the quadratic equation:

\(0.00012(t^2 - 100t + 2400) = 0.00012(t - 40)(t - 60) = 0\).

Thus, \(t = 40\) and \(t = 60\).

(ii) To find the displacement, integrate the velocity function from \(t = 0\) to \(t = 100\):

\(\int_0^{100} (0.00004t^3 - 0.006t^2 + 0.288t) \, dt\).

Calculate the integral:

\(\left[ 0.00001t^4 - 0.002t^3 + 0.144t^2 \right]_0^{100}\).

Evaluate at the limits:

\(0.00001(100)^4 - 0.002(100)^3 + 0.144(100)^2 = 440\).

Therefore, the displacement is 440 m.