(i) Let the speed of Q at time t be v_Q. Then, using the equation of motion,

\(v_Q = 3 + 2t\).

The speed of P is 1.8 times the speed of Q, so

\(v_P = 1.8v_Q = 1.8(3 + 2t) = 5 + 4t\).

Equating the expressions for \(v_P\), we have:

\(5 + 4t = 1.8(3 + 2t)\).

Solving for \(t\):

\(5 + 4t = 5.4 + 3.6t\)

\(0.4t = 0.4\)

\(t = 1\).

Substitute \(t = 1\) into \(v_P = 5 + 4t\):

\(v_P = 5 + 4(1) = 9 \text{ m s}^{-1}\).

(ii) Using the equation \(s = ut + \frac{1}{2}at^2\) for both particles:

For P: \(s_P = 5t + \frac{1}{2} \times 4 \times t^2\)

For Q: \(s_Q = 3t + \frac{1}{2} \times 2 \times t^2\)

Since \(s_P + s_Q = 51\):

\(5t + 2t^2 + 3t + t^2 = 51\)

\(3t^2 + 8t - 51 = 0\).

Solving the quadratic equation:

\((3t + 17)(t - 3) = 0\)

\(t = 3 \text{ s}\) (since time cannot be negative).

(a) For Isabella, the acceleration phase gives a final speed \(v = 5 \times 1.1 = 5.5 \text{ m s}^{-1}\). The total distance \(s_I\) is calculated as:

\(s_I = \frac{1}{2} \times 1.1 \times 5^2 + 10 \times 5.5 + \frac{1}{2} \times 1.1 \times 5^2 = 82.5 \text{ m}\)

For Maria, the total distance \(s_M\) is:

\(s_M = 27.5 + 5 \times 10 + \frac{1}{2} \times 5 \times 5 = 90 \text{ m}\)

Since \(s_M = 90\) and \(s_I = 82.5\), Maria travels further.

(b) To find \(a\) such that both travel the same distance, set \(s_I = s_M = 90\):

\(\frac{1}{2} a \times 5^2 + 10 \times 5a + \frac{1}{2} a \times 5^2 = 90\)

\(\frac{1}{2} a \times 25 + 50a + \frac{1}{2} a \times 25 = 90\)

\(25a + 50a = 90\)

\(75a = 90\)

\(a = 1.2\)

(i) The distance travelled by A is the area under the velocity-time graph. The graph consists of three segments: a triangle, a trapezium, and another triangle.

For the first triangle (0 to 100 s):

Area = \(\frac{1}{2} \times 100 \times 4.8 = 240\) m

For the trapezium (100 to 300 s):

Area = \(\frac{1}{2} \times 200 \times (4.8 + 7.2) = 1200\) m

For the second triangle (300 to 500 s):

Area = \(\frac{1}{2} \times 200 \times 7.2 = 720\) m

\(Total distance = 240 + 1200 + 720 = 2160 m\)

(ii) The initial acceleration of A is the gradient of the first line segment of the graph.

Acceleration \(a = \frac{4.8}{100} = 0.048\) m s^-2

(iii) The initial acceleration of B is given by differentiating the velocity function \(v = 0.06t - 0.00012t^2\).

\(a = \frac{dv}{dt} = 0.06 - 0.00024t\)

At \(t = 0\), \(a = 0.06\) m s^-2

The difference in initial acceleration is \(0.06 - 0.048 = 0.012\) m s^-2

(iv) B's velocity is maximum when \(0.06 - 0.00024t = 0\).

Solving gives \(t = 250\) s.

The distance travelled by B at \(t = 250\) s is found by integrating the velocity function:

\(s_B = \int (0.06t - 0.00012t^2) \, dt = 0.03t^2 - 0.00004t^3\)

\(s_B(250) = 0.03 \times 250^2 - 0.00004 \times 250^3 = 1095\) m

The distance travelled by A in the same time is 940 m (from the graph).

The difference is \(1095 - 940 = 155\) m

(i) To find the distance \(OP\), integrate the velocity function:

\(v = 0.6t^2 - 0.12t^3\).

Set \(v = 0\) to find when the particle comes to rest:

\(0.6t^2 - 0.12t^3 = 0\)

\(t^2(0.6 - 0.12t) = 0\)

\(t = 0\) or \(t = 5\).

Integrate \(v\) to find \(s\):

\(s = \int (0.6t^2 - 0.12t^3) \, dt = 0.2t^3 - 0.03t^4 + C\).

Since the particle starts from rest at \(O\), \(C = 0\).

Evaluate from \(t = 0\) to \(t = 5\):

\(OP = [0.2 \times 5^3 - 0.03 \times 5^4] - [0] = 6.25\) m.

(ii) Given \(s = kt^3 + ct^5\) and \(s = 6.25\) at \(t = 5\):

\(k \times 5^3 + c \times 5^5 = 6.25\).

Differentiate \(s\) to find \(v\):

\(v = 3kt^2 + 5ct^4\).

Given \(v = 1.25\) at \(t = 5\):

\(1.25 = 3k \times 5^2 + 5c \times 5^4\).

Solve the simultaneous equations:

\(125k + 3125c = 6.25\)

\(75k + 3125c = 1.25\).

Solving gives \(k = 0.1, c = -0.002\).

(iii) Differentiate \(v = 0.6t^2 - 0.12t^3\) to find acceleration \(a\):

\(a = \frac{dv}{dt} = 1.2t - 0.36t^2\).

At \(t = 5\), \(a = 1.2 \times 5 - 0.36 \times 5^2 = -2\) m s^-2.

(i) The velocity-time graph for P is a quadratic curve \(v = 6t(t-3) = 6t^2 - 18t\), which is a parabola opening upwards, passing through the origin \((0,0)\) and cutting the t-axis at \(t = 3\). The velocity-time graph for Q is a straight line \(v = 10 - 2t\), with a negative gradient, starting at \(v = 10\) when \(t = 0\) and reaching \(v = 0\) at \(t = 5\).

(ii) To verify that P and Q meet after 5 seconds, we find the displacement of each particle. For Q, the displacement is \(s = \int (10 - 2t) \, dt = 10t - t^2 + c\). Evaluating from \(t = 0\) to \(t = 5\), \(s(Q) = 25\) m. For P, the displacement is \(s = \int (6t^2 - 18t) \, dt = 2t^3 - 9t^2 + c\). Evaluating from \(t = 0\) to \(t = 5\), \(s(P) = 25\) m. Since both displacements are equal, P and Q meet at \(t = 5\).

(iii) The distance between P and Q is given by \(|s_P - s_Q| = |2t^3 - 8t^2 - 10t|\). To find the maximum distance, differentiate and solve \(6t^2 - 16t - 10 = 0\). Solving gives \(t = 3.19\). Evaluating the distance at this time gives the maximum distance \(PQ = 48.4\) m.