(i) For 0 ≤ t ≤ 10, the particle moves with constant acceleration a. At t = 10, the velocity is given by the velocity function: \(v = \frac{800}{10^2} - 2 = 8 - 2 = 6\) m s^-1. Therefore, the speed of P when t = 10 is 6 m s^-1. Since the particle starts from rest, \(v = at\), so \(a = \frac{6}{10} = 0.6\) m s^-2.

(ii) For 10 ≤ t ≤ 20, the acceleration is given by differentiating the velocity function: \(a(t) = \frac{d}{dt} \left( \frac{800}{t^2} - 2 \right) = -\frac{1600}{t^3}\). Set \(a(t) = -0.6\) to find \(t\): \(-\frac{1600}{t^3} = -0.6\) implies \(t^3 = \frac{1600}{0.6}\), so \(t = 13.9\).

(iii) To find the displacement from A when t = 20, integrate the velocity function: \(s = \int (\frac{800}{t^2} - 2) \, dt = -\frac{800}{t} - 2t + C\). Using \(s(10) = 30\) to find \(C\), we have \(-\frac{800}{10} - 20 + C = 30\), so \(C = 130\). Then, \(s(20) = -\frac{800}{20} - 40 + 130 = 50\). Therefore, the displacement from A is 50 m.

(a) To find the velocity at \(t = 4\), integrate the acceleration \(a = 0.3t^{\frac{1}{2}}\) from \(t = 0\) to \(t = 4\):

\(v = \int 0.3t^{\frac{1}{2}} \, dt = 0.3 \cdot \frac{2}{3} t^{\frac{3}{2}} + c = 0.2t^{\frac{3}{2}} + c\).

Since the particle starts from rest, \(c = 0\). Thus, \(v = 0.2 \times 4^{\frac{3}{2}} = 1.6 \text{ m/s}\).

(b) For \(4 < t \leq T\), \(a = -kt^{-\frac{3}{2}}\). Integrate to find velocity:

\(v = \int -kt^{-\frac{3}{2}} \, dt = -k \cdot \frac{-2}{1} t^{-\frac{1}{2}} + d = 2kt^{-\frac{1}{2}} + d\).

Given \(v = 0.3 \text{ m/s}\) at \(t = 16\), and no change at \(t = 4\), equate velocities:

\(1.6 = 2k \cdot 4^{-\frac{1}{2}} + d\) and \(0.3 = 2k \cdot 16^{-\frac{1}{2}} + d\).

Solving gives \(k = 2.6\) and \(v = 5.2t^{-\frac{1}{2}} - 1\).

(c) Set \(v = 0\) for \(t = T\):

\(5.2T^{-\frac{1}{2}} - 1 = 0\).

Solving gives \(T = \frac{676}{25}\) or 27.04.

(d) Total distance is the integral of speed from \(t = 0\) to \(t = T\):

\(\int_0^4 0.2t^{\frac{1}{2}} \, dt + \int_4^{27.04} (5.2t^{-\frac{1}{2}} - 1) \, dt\).

Calculating gives 12.8 m.

(i) The acceleration \(a\) is given by \(a = \frac{dv}{dt} = 0.5 - 0.02t\). Setting \(a = 0.1\), we solve \(0.5 - 0.02t = 0.1\) to find \(t = 20\) s.

(ii) The initial velocity at B is \(u = 0.5 \times 20 - 0.01 \times 20^2 = 6\) m s\(^{-1}\). Using \(v = u + at\), where \(v = 14\) m s\(^{-1}\) and \(a = 0.1\) m s\(^{-2}\), we have \(14 = 6 + 0.1t\). Solving gives \(t = 80\) s.

(iii) Using \(v^2 = u^2 + 2as\) for the section CD, where \(u = 14\) m s\(^{-1}\), \(a = -0.3\) m s\(^{-2}\), and \(s = 300\) m, we find \(v^2 = 14^2 - 2 \times 0.3 \times 300\). Solving gives \(v = 4\) m s\(^{-1}\).

(iv) The displacement \(s\) is given by integrating the velocity function: \(s = \int v \, dt = 0.25t^2 - 0.01t^3/3 + C\). For AB, \(s = 0.25 \times 20^2 - 0.01 \times 20^3/3 = 73.3\) m. For BC, \(s = \frac{1}{2}(6 + 14) \times 80 = 800\) m. Thus, the total distance AD is \(73.3 + 800 + 300 = 1170\) m.

(i) To find the maximum velocity, we need to find the maximum point of the curve v = -0.01t² + 0.5t - 1 for 10 ≤ t ≤ 30. Differentiate v with respect to t:

\(\frac{dv}{dt} = -0.02t + 0.5\)

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

\(-0.02t + 0.5 = 0\)

\(t = 25\)

Substitute \(t = 25\) back into the velocity equation:

\(v = -0.01(25)^2 + 0.5(25) - 1 = 5.25\)

Thus, the maximum velocity is 5.25 m s^-1.

(ii) To find the distance AB, calculate the area under the velocity-time graph. The graph consists of three parts:

• From 0 to 10 s: A straight line with velocity increasing from 0 to 3 m s^-1. The area is a triangle: \(s_1 = \frac{1}{2} \times 3 \times 10 = 15\) m.
• From 10 to 30 s: The curved section. Integrate the velocity function:

\(s_2 = \int_{10}^{30} (-0.01t^2 + 0.5t - 1) \, dt\)

\(s_2 = \left[ -0.01 \frac{t^3}{3} + 0.5 \frac{t^2}{2} - t \right]_{10}^{30}\)

\(s_2 = \left[ -0.01 \times \frac{30^3}{3} + 0.5 \times \frac{30^2}{2} - 30 \right] - \left[ -0.01 \times \frac{10^3}{3} + 0.5 \times \frac{10^2}{2} - 10 \right]\)

\(s_2 = ( -90 + 225 - 30 ) - ( -10/3 + 25 - 10 ) = 93.3\) m

• From 30 to 80 s: A straight line with velocity decreasing from 5 to 0 m s^-1. The area is a triangle: \(s_3 = \frac{1}{2} \times 5 \times 50 = 125\) m.

Total distance \(AB = s_1 + s_2 + s_3 = 15 + 93.3 + 125 = 233.3\) m.

(i) The acceleration \(a(t)\) is the derivative of the velocity \(v(t)\) with respect to time \(t\). Thus,

\(a(t) = \frac{dv}{dt} = \frac{d}{dt}(1.25t - 0.05t^2) = 1.25 - 0.1t\).

The initial acceleration is when \(t = 0\):

\(a(0) = 1.25 - 0.1 \times 0 = 1.25 \text{ m/s}^2\).

(ii) To find the displacement when the acceleration is \(0.05 \text{ m/s}^2\), set \(a(t) = 0.05\):

\(1.25 - 0.1t = 0.05\)

\(0.1t = 1.25 - 0.05\)

\(0.1t = 1.2\)

\(t = 12 \text{ s}\).

Now, integrate \(v(t)\) to find the displacement \(s(t)\):

\(s(t) = \int v(t) \, dt = \int (1.25t - 0.05t^2) \, dt\)

\(= \frac{1.25t^2}{2} - \frac{0.05t^3}{3} + C\).

Since the particle starts from rest, \(s(0) = 0\), so \(C = 0\).

Calculate \(s(12)\):

\(s(12) = \frac{1.25 \times 12^2}{2} - \frac{0.05 \times 12^3}{3}\)

\(= \frac{1.25 \times 144}{2} - \frac{0.05 \times 1728}{3}\)

\(= 90 - 28.8\)

\(= 61.2 \text{ m}\).