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

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

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

\(k = 10\).

\((i) The particle is decelerating when the acceleration is negative. Given a = 6t - 2, set 6t - 2 < 0:\)

\(6t - 2 < 0\)

\(6t < 2\)

\(t < \frac{1}{3}\)

\(Thus, the particle is decelerating for 0 < t < \frac{1}{3}.\)

(ii) To find s in terms of t, first find the velocity v by integrating the acceleration:

\(v = \int (6t - 2) \, dt = 3t^2 - 2t + c\)

Next, integrate the velocity to find the displacement s:

\(s = \int (3t^2 - 2t + c) \, dt = t^3 - t^2 + ct + d\)

\(Using the conditions t = 1, s = 7 and t = 3, s = 29, form the equations:\)

\(c + d = 7\)

\(3c + d = 11\)

\(Solving these simultaneous equations gives c = 2 and d = 5.\)

\(Thus, s = t^3 - t^2 + 2t + 5.\)

\((iii) To find the time when the velocity is 10 m s^-1, set v = 10:\)

\(3t^2 - 2t + 2 = 10\)

\(3t^2 - 2t - 8 = 0\)

\(Solving this quadratic equation gives t = 2.\)

(i) To find when P is at instantaneous rest, set the velocity v to zero:

\(4t^2 - 8t + 3 = 0\)

Factor the quadratic equation:

\((2t - 3)(2t - 1) = 0\)

Solving gives \(t = 0.5\) and \(t = 1.5\).

(ii) To find the distance travelled, integrate the velocity function:

\(s = -\int (4t^2 - 8t + 3) \, dt\)

Integrate to find the displacement:

\(s = -\left[ \frac{4}{3}t^3 - 4t^2 + 3t \right]_{0.5}^{1.5}\)

Evaluate the definite integral:

\(s = -\left[ \left( \frac{4}{3}(1.5)^3 - 4(1.5)^2 + 3(1.5) \right) - \left( \frac{4}{3}(0.5)^3 - 4(0.5)^2 + 3(0.5) \right) \right]\)

\(s = -\left[ 0 - \left( -\frac{2}{3} \right) \right] = \frac{2}{3}\) m

(a) We have the equations from the given conditions:

\(4b + 4^{\frac{3}{2}}c = 8\)

\(9b + 9^{\frac{3}{2}}c = 13.5\)

Solving these simultaneously, we find \(b = 3\) and \(c = -0.5\).

(b) The acceleration \(a\) is given by \(a = \frac{dv}{dt} = b + \frac{3}{2}ct^{\frac{1}{2}}\).

Substituting \(b = 3\), \(c = -0.5\), and \(t = 1\), we get:

\(a = 3 + \frac{3}{2}(-0.5)(1)^{\frac{1}{2}} = 3 - 0.75 = 2.25 \text{ m/s}^2\)

(c) For instantaneous rest, set \(v = 0\):

\(3t - 0.5t^{\frac{3}{2}} = 0\)

Solving gives \(t = 36\).

To find the distance, integrate \(v\) to get \(s\):

\(s = \int (3t - 0.5t^{\frac{3}{2}}) \, dt = \frac{3}{2}t^2 - \frac{1}{5}t^{\frac{5}{2}} + C\)

Using limits from 0 to 36, \(s = 388.8 \text{ m}\).

(d) For the speed when returning to \(O\), solve:

\(\frac{3}{2}t^2 - \frac{1}{5}t^{\frac{5}{2}} = 0\)

\(t = 56.25\).

Substitute back into \(v\):

\(v = 3(56.25) - 0.5(56.25)^{\frac{3}{2}} = 42.2 \text{ m/s}\)

(i) The acceleration \(a\) is the derivative of the velocity \(v\). So, \(a = \frac{dv}{dt} = 12t - 30\). For the acceleration to be negative, \(12t - 30 < 0\). Solving this inequality gives \(t < 2.5\).

(ii) The particle is at instantaneous rest when \(v = 0\). Solving \(6t^2 - 30t + 24 = 0\) gives \(t = 1\) and \(t = 4\). The distance \(s\) is the integral of \(v\), so \(s = \int (6t^2 - 30t + 24) \, dt = 2t^3 - 15t^2 + 24t\). Evaluating from \(t = 1\) to \(t = 4\), \(s = [2t^3 - 15t^2 + 24t]_1^4\), which gives a distance of 27 m.

(iii) The particle passes through \(O\) when \(s = 0\). Solving \(2t^3 - 15t^2 + 24t = 0\) gives \(t = 0\), \(t = 2.31\), and \(t = 5.19\). The positive values are \(t = 2.31\) and \(t = 5.19\).