(i) To find the maximum acceleration, differentiate \(a = 15t - 3t^2\) with respect to \(t\):

\(\frac{da}{dt} = 15 - 6t\)

Set \(\frac{da}{dt} = 0\) to find critical points:

\(15 - 6t = 0\)

\(t = 2.5\) seconds.

Substitute \(t = 2.5\) into \(a = 15t - 3t^2\):

\(a = 15(2.5) - 3(2.5)^2 = 18.75\) m s\(^{-2}\).

(ii) To find the distance, first find the velocity by integrating the acceleration:

\(v(t) = \int (15t - 3t^2) \, dt = 7.5t^2 - t^3 + c\)

Since the car starts from rest, \(v(0) = 0\), so \(c = 0\).

Now integrate the velocity to find the distance:

\(s(t) = \int (7.5t^2 - t^3) \, dt = 2.5t^3 - 0.25t^4 + d\)

Since \(s(0) = 0\), \(d = 0\).

At \(t = 5\),

\(s(5) = 2.5(5)^3 - 0.25(5)^4 = 156.25\) m.

(iii) The car comes to rest when \(v(k) = 0\). First, find \(v(5)\):

\(v(5) = 7.5(5)^2 - (5)^3 = 62.5\) m s\(^{-1}\).

For \(5 < t \leq k\), integrate \(a = -\frac{625}{t^2}\) to find \(v(t)\):

\(v(t) = \int -\frac{625}{t^2} \, dt = \frac{625}{t} + c\)

Using \(v(5) = 62.5\),

\(62.5 = \frac{625}{5} + c\)

\(c = -62.5\).

Set \(v(k) = 0\):

\(\frac{625}{k} - 62.5 = 0\)

\(\frac{625}{k} = 62.5\)

\(k = 10\).