(i) To find the distance travelled in the first 8 seconds, calculate the area under the velocity-time graph from \(t = 0\) to \(t = 8\).

The velocity at \(t = 3\) is \(3 \times 3 = 9 \text{ m s}^{-1}\).

The area under the graph from \(t = 0\) to \(t = 8\) consists of a triangle and a trapezium:

Area of triangle from \(t = 0\) to \(t = 3\): \(\frac{1}{2} \times 3 \times 9 = 13.5 \text{ m}\).

Area of trapezium from \(t = 3\) to \(t = 8\): \(\frac{1}{2} \times (9 + 7) \times 2 + \frac{1}{2} \times 3 \times 7 = 26.5 \text{ m}\).

Total distance = \(13.5 + 26.5 = 40 \text{ m}\).

(ii) The total displacement from \(t = 0\) to \(t = 16\) is given as 32 m. The distance from \(t = 0\) to \(t = 8\) is 40 m, so the displacement from \(t = 8\) to \(t = 16\) must be \(32 - 40 = -8 \text{ m}\).

The area under the graph from \(t = 8\) to \(t = 16\) is a triangle with base \(8\) and height \(V\).

Area of triangle = \(\frac{1}{2} \times 8 \times V = -8\).

Solving for \(V\), we get \(V = -2\).

(i) The acceleration during the first 2 seconds is the gradient of the velocity-time graph from \(t = 0\) to \(t = 2\). The change in velocity is from \(0\) to \(-2\) m s^-1, so the acceleration \(a\) is given by:

\(a = \frac{-2 - 0}{2 - 0} = -1 \text{ m s}^{-2}\).

(ii) To find \(V\), use the gradient of the line between \(t = 4\) and \(t = 10\). The change in velocity is from \(-2\) to \(V\), and the change in time is \(6\) seconds:

\(\frac{V + 2}{6} = 1\).

Solving for \(V\), we get:

\(V + 2 = 6\)

\(V = 4\).

(iii) The distance \(AB\) is three times the distance traveled in the first 6 seconds. The area under the graph from \(t = 0\) to \(t = 6\) is:

\(\frac{1}{2} \times (6 + 2) \times 2 = 8 \text{ m}\).

Thus, \(AB = 3 \times 8 = 24 \text{ m}\).

For \(t = T\), the area under the graph from \(t = 6\) to \(t = T\) must also be \(16 \text{ m}\) (since \(AB = 24 \text{ m}\) and \(8 \text{ m}\) is already covered):

\(\frac{1}{2} \times (T - 6) \times 4 = 16\).

Solving for \(T\), we get:

\(2(T - 6) = 16\)

\(T - 6 = 8\)

\(T = 18\).

(i) The acceleration is calculated using the formula:

\(a = \frac{v - u}{t}\)

where \(v = 16 \text{ m/s}\), \(u = 0 \text{ m/s}\), and \(t = 40 \text{ s}\).

Thus, \(a = \frac{16 - 0}{40} = 0.4 \text{ m/s}^2\).

(ii) The total distance is the area under the velocity-time graph, which is a trapezium. The area \(A\) is given by:

\(A = \frac{1}{2} \times (600 + T) \times 16 = 9040\)

Solving for \(T\):

\(9040 = \frac{1}{2} \times (600 + T) \times 16\)

\(9040 = 8 \times (600 + T)\)

\(1130 = 600 + T\)

\(T = 530 \text{ s}\)

(iii) The distance travelled while decelerating is the area of the triangle at the end of the graph:

\(s = \frac{1}{2} \times (600 - 530 - 40) \times 16\)

\(s = \frac{1}{2} \times 30 \times 16\)

\(s = 240 \text{ m}\)

(i) The velocity-time graph consists of three straight line segments: a positive gradient for 0 < t < 80, a zero gradient for 80 < t < 560, and a negative gradient for 560 < t < 600. The graph starts and ends on the t-axis at t = 0 and t = 600, respectively.

(ii) Using the area under the velocity-time graph for 0 < t < 80, we have:

\(\frac{1}{2} \times 80 \times v = 400\)

Solving for \(v\), we get \(v = 10 \text{ ms}^{-1}\).

(iii) The total distance is the area under the velocity-time graph. The area of the trapezium is:

\(D = \frac{1}{2} (600 + 480) \times 10 = 5400 \text{ m}\)

(iv) The acceleration for 0 < t < 80 is given by the gradient of the velocity-time graph:

\(v = 0 + at\)

\(10 = a \times 80\)

\(a = 0.125 \text{ ms}^{-2}\)

(a) For particle A, it moves with constant speed 8 m/s. Since B passes O 4 seconds after A, the displacement of A after t seconds is:

\(s_A = 8(4 + t) = 32 + 8t\)

For particle B, using the equation \(s = ut + \frac{1}{2}at^2\) with \(u = 20\) m/s and \(a = -2\) m/s², the displacement is:

\(s_B = 20t - \frac{1}{2} \times 2t^2 = 20t - t^2\)

(b) Set \(s_A = s_B\):

\(32 + 8t = 20t - t^2\)

Rearrange to form a quadratic equation:

\(t^2 - 12t + 32 = 0\)

Solving this gives \(t = 4\) and \(t = 8\).

(c) The graph for A is a straight line starting at \(s = 32\) when \(t = 0\) with a positive gradient. The graph for B is an inverted quadratic starting at the origin, reaching a maximum, and then returning to the axis. They intersect at \(t = 4\) and \(t = 8\).