(i) Using the equation of motion: \(v = u + at\), where \(v = 5\) m/s, \(u = 2\) m/s, and \(a = 0.05\) m/s², we have:

\(5 = 2 + 0.05t\)

\(0.05t = 3\)

\(t = \frac{3}{0.05} = 60\) s

Using \(v^2 = u^2 + 2as\), we find the distance \(s\):

\(5^2 = 2^2 + 2 \times 0.05 \times s\)

\(25 = 4 + 0.1s\)

\(0.1s = 21\)

\(s = \frac{21}{0.1} = 210\) m

(ii) For particle Q, the speed is given by \(v = kt^3\). The distance \(s\) is given by integrating the speed:

\(s = \int_0^{60} kt^3 \, dt = \left[ \frac{kt^4}{4} \right]_0^{60} = \frac{k \times 60^4}{4}\)

Since \(s = 210\) m, we have:

\(210 = \frac{k \times 60^4}{4}\)

\(k = \frac{210 \times 4}{60^4} = \frac{7}{108000}\)

The speed of Q at B is \(v = k \times 60^3\):

\(v = \frac{7}{108000} \times 60^3 = 14\) m/s

(i) Using the equation of motion:

\(v = u + at\)

where \(v = 1.3\) m/s, \(u = 0.9\) m/s, and \(a = 0.004\) m/s², we have:

\(1.3 = 0.9 + 0.004T\)

Solving for \(T\):

\(T = \frac{1.3 - 0.9}{0.004} = 100 \text{ s}\)

Using the equation:

\(v^2 = u^2 + 2as\)

where \(s\) is the distance, we have:

\(1.3^2 = 0.9^2 + 2 \times 0.004 \times s\)

Solving for \(s\):

\(s = \frac{1.3^2 - 0.9^2}{2 \times 0.004} = 110 \text{ m}\)

(ii) The cyclist's speed is given by:

\(v_c = kt^3\)

Integrating to find the distance:

\(\int kt^3 \, dt = \frac{1}{4} kt^4\)

Setting the limits from 0 to 100:

\(\frac{1}{4} k (100)^4 = 110\)

Solving for \(k\):

\(k = \frac{110 \times 4}{100^4} = 4.4 \times 10^{-6}\)

At \(t = 64.05\):

\(v_w = 0.9 + 0.004 \times 64.05\)

\(v_c = 4.4 \times 10^{-6} \times (64.05)^3\)

Both speeds are:

\(v_w = v_c = 1.16 \text{ m/s}\)

(iii) The cyclist's acceleration is:

\(a_c = \frac{d}{dt}(kt^3) = 3kt^2\)

At \(t = 100\):

\(a_c = 3 \times 4.4 \times 10^{-6} \times 100^2 = 0.132 \text{ m/s}^2\)

\((i) To find the displacement of P from O when t = 10, we use the formula for displacement with constant acceleration:\)

\(s = \frac{(u + v)}{2} \times t\)

where \(u = 1.5\) m/s, \(v = 3.5\) m/s, and \(t = 10\) s.

\(s = \frac{(1.5 + 3.5)}{2} \times 10 = 25 \text{ m}\)

(ii) For particle Q, the acceleration is given by \(a = 0.03t\). To find the velocity, integrate the acceleration:

\(v = \int 0.03t \, dt = 0.015t^2 + C\)

Given that \(v = 3.5\) m/s when \(t = 10\), we find \(C\):

\(3.5 = 0.015 \times 100 + C\)

\(3.5 = 1.5 + C\)

\(C = 2\)

Thus, the velocity equation is \(v = 0.015t^2 + 2\).

To find the displacement, integrate the velocity:

\(s = \int (0.015t^2 + 2) \, dt = 0.005t^3 + 2t + D\)

Since Q starts from O, \(D = 0\).

At \(t = 10\):

\(s = 0.005 \times 1000 + 20 = 5 + 20 = 25 \text{ m}\)

Therefore, the displacement of Q is also 25 m, the same as P.

(i) Using the equation of motion:

\(v = u + at\)

where \(v = 2.2\) m s^-1, \(u = 1.8\) m s^-1, and \(a = 0.004\) m s^-2.

Substitute the values:

\(2.2 = 1.8 + 0.004t\)

Solve for \(t\):

\(0.004t = 0.4\)

\(t = 100 \text{ s}\)

To find the distance \(s\), use:

\(s = ut + \frac{1}{2}at^2\)

Substitute the values:

\(s = 1.8 \times 100 + \frac{1}{2} \times 0.004 \times 100^2\)

\(s = 180 + 20 = 200 \text{ m}\)

(ii) (a) Integrate the velocity function to find the displacement \(s(t)\):

\(s = \int k(200t - t^2) \, dt = k \left( 100t^2 - \frac{t^3}{3} \right) + C\)

Given \(s(0) = 0\), \(C = 0\).

At \(t = 100\), \(s(100) = 200\):

\(k \left( 100 \times 100^2 - \frac{100^3}{3} \right) = 200\)

Solve for \(k\):

\(k \times 666666.67 = 200\)

\(k = 0.0003\)

(b) The cyclist’s speed at \(t = 100\) is:

\(v = k(200 \times 100 - 100^2)\)

\(v = 0.0003 \times 10000 = 3 \text{ m s}^{-1}\)

(iii) The velocity-time graph for the man is a straight line with positive slope, starting at 1.8 m s^-1 and ending at 2.2 m s^-1. The cyclist’s graph is a parabola with decreasing slope, starting at 0 m s^-1 and ending at 3 m s^-1.

(i) The displacement of Q is given by integrating the velocity function:

\(s_Q = \int v_Q \, dt = \int (3t - 0.3t^2) \, dt = 1.5t^2 - 0.1t^3 + C\).

Using the limits from 0 to 10, we find:

\(s_Q(10) = 1.5(10)^2 - 0.1(10)^3 = 150 - 100 = 50\).

The displacement of P is the area under the velocity-time graph, which is a triangle with base 10 and height 10:

\(s_P = \frac{1}{2} \times 10 \times 10 = 50\).

Since \(s_P = s_Q\), the greatest velocity of P is 10 m s^-1.

(ii) The acceleration of P is constant during the interval \(0 < t < 5\) and is given by:

\(a_P = \frac{10}{5} = 2 \text{ m s}^{-2}\).

The acceleration of Q is found by differentiating the velocity function:

\(a_Q = \frac{d}{dt}(3t - 0.3t^2) = 3 - 0.6t\).

Setting \(a_Q = a_P\):

\(3 - 0.6t = 2\).

Solving for \(t\):

\(0.6t = 1\).

\(t = \frac{1}{0.6} = 1.67\) (or \(1\frac{2}{3}\)).