(i) For \(t = 3\), substitute into the velocity equation: \(v = -0.2(3)^2 + 4(3) - 15 = -4.8\) m s^-1. The acceleration for the first 3 s is constant, using \(v = u + at\) with \(u = 0\), \(v = -4.8\), and \(t = 3\), gives \(a = \frac{-4.8}{3} = -1.6\) m s^-2.

(ii) To find the maximum velocity, set \(\frac{dv}{dt} = 0\). Differentiate \(v = -0.2t^2 + 4t - 15\) to get \(\frac{dv}{dt} = -0.4t + 4\). Solve \(-0.4t + 4 = 0\) to find \(t = 10\). Substitute \(t = 10\) into the velocity equation: \(v = -0.2(10)^2 + 4(10) - 15 = 5\) m s^-1.

(iii) (a) Distance from 0 to 3 s is \(\frac{1}{2} \times 3 \times 4.8 = 7.2\) m. From 3 to 5 s, integrate \(v = -0.2t^2 + 4t - 15\) from 3 to 5: \(\int_{3}^{5} (-0.2t^2 + 4t - 15) \, dt = 4.533\) m. Total distance = 7.2 + 4.533 = 11.733 m. Average speed = \(\frac{11.733}{5} = 2.35\) m s^-1.

(b) For the whole journey, integrate from 3 to 15: \(\int_{3}^{15} (-0.2t^2 + 4t - 15) \, dt = 45.066\) m. Total distance = 7.2 + 45.066 = 52.266 m. Average speed = \(\frac{52.266}{15} = 3.00\) m s^-1.

(i) For \(t > 8\), the velocity \(v(t) = \frac{1}{2} t^{2/3}\). The acceleration \(a(t)\) is the derivative of \(v(t)\) with respect to \(t\):

\(a(t) = \frac{d}{dt} \left( \frac{1}{2} t^{2/3} \right) = \frac{1}{3} t^{-1/3}\).

At \(t = 8\), the acceleration just before passing \(A\) is \(\frac{1}{4} \text{ m s}^{-2}\) and just after is \(\frac{1}{3} \times 8^{-1/3} = \frac{1}{6} \text{ m s}^{-2}\).

The decrease in acceleration is \(\frac{1}{4} - \frac{1}{6} = \frac{1}{12} \text{ m s}^{-2}\).

(ii) The distance moved from \(t = 0\) to \(t = 8\) is given by:

\(s_1 = \frac{1}{2} \times \frac{1}{4} \times 8^2 = 8 \text{ m}\).

For \(t > 8\), the distance \(s_2\) is:

\(s_2 = \int_{8}^{27} \frac{1}{2} t^{2/3} \, dt = \left[ 0.3 t^{5/3} \right]_{8}^{27}\).

Calculating the definite integral:

\(s_2 = 0.3 \times 27^{5/3} - 0.3 \times 8^{5/3} = 71.3 - 8 = 63.3 \text{ m}\).

Total distance \(s = s_1 + s_2 = 8 + 63.3 = 71.3 \text{ m}\).

(i) To find \(k_1\), use the formula for distance: \(s = \int v \, dt\). For \(0 \leq t \leq 60\),

\(s = \int_0^{60} (k_1 t - 0.005t^2) \, dt = \left[ \frac{k_1 t^2}{2} - 0.005 \frac{t^3}{3} \right]_0^{60} = 540\)

\(k_1 \frac{60^2}{2} - 0.005 \frac{60^3}{3} = 540\)

\(1800k_1 - 72 = 540\)

\(1800k_1 = 612\)

\(k_1 = 0.5\)

For \(k_2\), equate \(v(60)\) from both expressions:

\(v(60) = 0.5 \times 60 - 0.005 \times 60^2 = \frac{k_2}{\sqrt{60}}\)

\(30 - 18 = \frac{k_2}{\sqrt{60}}\)

\(12 = \frac{k_2}{\sqrt{60}}\)

\(k_2 = 12\sqrt{60}\)

(ii) For \(t \geq 60\), the distance is:

\(s = 540 + \int_{60}^{t} \frac{12\sqrt{60}}{\sqrt{t}} \, dt\)

\(s = 540 + 12\sqrt{60} \left[ 2\sqrt{t} - 2\sqrt{60} \right]\)

\(s = 24\sqrt{60t} - 900\)

(iii) Set \(s(t) = 1260\):

\(24\sqrt{60t} - 900 = 1260\)

\(24\sqrt{60t} = 2160\)

\(\sqrt{60t} = 90\)

\(60t = 8100\)

\(t = 135\)

Find speed at \(t = 135\):

\(v = \frac{12\sqrt{60}}{\sqrt{135}} = 8 \text{ m s}^{-1}\)

(i) For 0 ≤ t ≤ 5, the velocity is given by v = t - 0.1t². The distance travelled, s_1, is the integral of velocity with respect to time:

\(s_1 = \int_0^5 (t - 0.1t^2) \, dt = \left[ \frac{t^2}{2} - \frac{0.1t^3}{3} \right]_0^5\)

\(s_1 = \left( \frac{25}{2} - \frac{0.1 \times 125}{3} \right) = 8.33 \text{ m}\)

(ii) For 5 ≤ t ≤ 45, the velocity is constant. Let the constant velocity be v_c. Since the particle starts from rest and returns to rest, the symmetry implies v_c = 2.5 m/s. The distance travelled, s_2, is:

\(s_2 = v_c \times (45 - 5) = 2.5 \times 40 = 100 \text{ m}\)

\(For 45 ≤ t ≤ 50, the velocity is given by v = 9t - 0.1t^2 - 200. The distance travelled, s_3, is:\)

\(s_3 = \int_{45}^{50} (9t - 0.1t^2 - 200) \, dt\)

\(s_3 = \left[ \frac{9t^2}{2} - \frac{0.1t^3}{3} - 200t \right]_{45}^{50}\)

\(s_3 = \left( \frac{9 \times 2500}{2} - \frac{0.1 \times 125000}{3} - 200 \times 50 \right) - \left( \frac{9 \times 2025}{2} - \frac{0.1 \times 91125}{3} - 200 \times 45 \right)\)

\(s_3 = 8.33 \text{ m}\)

The total distance from O to A is:

\(s = s_1 + s_2 + s_3 = 8.33 + 100 + 8.33 = 117 \text{ m}\)

The average speed is:

\(\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{117}{50} = 2.33 \text{ m/s}\)

To find when the particle is at rest, we need to find when the velocity \(v = \frac{ds}{dt}\) is zero.

Differentiate \(s = t^{\frac{5}{2}} - \frac{15}{4} t^{\frac{3}{2}} + 6\) with respect to \(t\):

\(v = \frac{5}{2} t^{\frac{3}{2}} - \frac{45}{8} t^{\frac{1}{2}}\).

Set \(v = 0\):

\(\frac{5}{2} t^{\frac{3}{2}} - \frac{45}{8} t^{\frac{1}{2}} = 0\).

Factor out \(t^{\frac{1}{2}}\):

\(t^{\frac{1}{2}} \left( \frac{5}{2} t - \frac{45}{8} \right) = 0\).

Since \(t^{\frac{1}{2}} = 0\) is not possible (as \(t = 0\) is the starting point), solve:

\(\frac{5}{2} t - \frac{45}{8} = 0\).

\(5t = \frac{45}{4}\).

\(t = \frac{45}{20} = 2.25\).

Substitute \(t = 2.25\) back into the displacement equation:

\(s = (2.25)^{\frac{5}{2}} - \frac{15}{4} (2.25)^{\frac{3}{2}} + 6\).

Calculate \(s\):

\(s = \frac{15}{16}\).