(a) To find when the particle is at instantaneous rest, set the velocity \(v = 0\). First, integrate the acceleration to find the velocity:

\(v = \int (36 - 6t) \, dt = 36t - 3t^2 + c\)

Given \(v = 27\) when \(t = 2\), substitute to find \(c\):

\(27 = 36(2) - 3(2)^2 + c\)

\(27 = 72 - 12 + c\)

\(c = -33\)

Thus, the velocity equation is:

\(v = 36t - 3t^2 - 33\)

Set \(v = 0\) to find \(t\):

\(0 = 36t - 3t^2 - 33\)

\(3t^2 - 36t + 33 = 0\)

Solving the quadratic equation gives \(t = 1\) and \(t = 11\).

(b) To find the total distance traveled, integrate the velocity to find the displacement:

\(s = \int (36t - 3t^2 - 33) \, dt = 18t^2 - t^3 - 33t + c\)

Evaluate from \(t = 0\) to \(t = 1\), \(t = 1\) to \(t = 11\), and \(t = 11\) to \(t = 12\):

\(s(0 \to 1) = 18(1)^2 - (1)^3 - 33(1) = -16\)

\(s(1 \to 11) = [18(11)^2 - (11)^3 - 33(11)] - [-16] = 500\)

\(s(11 \to 12) = [18(12)^2 - (12)^3 - 33(12)] - [18(11)^2 - (11)^3 - 33(11)] = -16\)

Total distance = \(|16| + 500 + |16| = 532 \text{ m}\).

(a) To verify that the particle returns to O when t = 2, we need to find the displacement from t = 0 to t = 2. The displacement is given by the integral of the velocity function:

\(\int_0^2 k(3t^2 - 2t^3) \, dt = k \left[ \frac{t^3}{3} - \frac{t^4}{4} \right]_0^2\)

\(= k \left( \frac{8}{3} - 4 \right) = k \left( \frac{8}{3} - \frac{12}{3} \right) = k \left( -\frac{4}{3} \right)\)

Since the displacement is zero, \(k \left( -\frac{4}{3} \right) = 0\), which implies \(k = 0\) or the displacement is zero, confirming the particle returns to O.

(b) The acceleration \(a\) is given by differentiating the velocity function:

\(a = \frac{dv}{dt} = k(6t - 6t^2)\)

Given \(a = -13.5\) m s^-2 when \(v = 0\), solve \(k(3t^2 - 2t^3) = 0\):

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

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

\(t = 0\) or \(t = 1.5\). For \(t = 1.5\), substitute into the acceleration equation:

\(-13.5 = k(6 \times 1.5 - 6 \times 1.5^2)\)

\(-13.5 = k(9 - 13.5)\)

\(-13.5 = -4.5k\)

\(k = 3\).

To find the total distance travelled in the first two seconds, calculate the integral of the absolute value of velocity from \(t = 0\) to \(t = 1.5\) and double it:

\(\int_0^{1.5} 3(3t^2 - 2t^3) \, dt = 3 \left[ \frac{t^3}{3} - \frac{t^4}{4} \right]_0^{1.5}\)

\(= 3 \left( \frac{1.5^3}{3} - \frac{1.5^4}{4} \right)\)

\(= 3 \left( \frac{3.375}{3} - \frac{5.0625}{4} \right)\)

\(= 3 \left( 1.125 - 1.265625 \right)\)

\(= 3(-0.140625) = -0.421875\)

Since distance is positive, the total distance is \(2 \times 2.53125 = 5.06\) m.

(a) To find the value of k, integrate the acceleration to find the velocity:

\(v = \int (2t - \frac{3}{5}t^2) \, dt = t^2 - \frac{1}{5}t^3 + C\)

Since the cyclist starts from rest, \(v = 0\) when \(t = 0\), so \(C = 0\).

Set \(v = 0\) to find \(k\):

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

\(t^2(1 - \frac{1}{5}t) = 0\)

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

Since \(t = 0\) is the starting point, \(k = 10\).

(b) The maximum speed occurs when acceleration \(a = 0\):

\(2t - \frac{3}{5}t^2 = 0\)

\(t(2 - \frac{3}{5}t) = 0\)

\(t = 0\) or \(t = \frac{10}{3}\)

Substitute \(t = \frac{10}{3}\) into the velocity equation:

\(v = (\frac{10}{3})^2 - \frac{1}{5}(\frac{10}{3})^3\)

\(v = \frac{100}{9} - \frac{1000}{135}\)

\(v = \frac{800}{135} = \frac{8\sqrt{30}}{9} \approx 4.87 \text{ m/s}\)

(c) Integrate the velocity to find displacement:

\(s = \int (t^2 - \frac{1}{5}t^3) \, dt = \frac{t^3}{3} - \frac{t^4}{20} + C\)

Substitute \(t = \frac{10}{3}\) and \(t = 10\) to find the distance:

\(s(\frac{10}{3}) = \frac{(\frac{10}{3})^3}{3} - \frac{(\frac{10}{3})^4}{20}\)

\(s(10) = \frac{10^3}{3} - \frac{10^4}{20}\)

Distance = \(s(10) - s(\frac{10}{3}) \approx 20.7 \text{ m}\)

(a) The acceleration is given by \(a = 16k - kt^2\). The velocity \(v\) is the integral of acceleration: \(v = \int a \, dt = 16kt - \frac{1}{3} kt^3\).

Given \(v = 8\) when \(t = 4\), substitute to find \(k\):

\(8 = 16k \times 4 - \frac{1}{3} k \times 4^3\)

Solve for \(k\): \(k = \frac{3}{16}\).

Now, find \(s\) by integrating \(v\):

\(s = \int v \, dt = 8kt^2 - \frac{1}{12} kt^4\)

Substitute \(k = \frac{3}{16}\):

\(s = \frac{3}{16} \times 8t^2 - \frac{3}{16} \times \frac{1}{12} t^4\)

\(s = \frac{1}{64} t^2 (96 - t^2)\)

(b) To find the speed when \(s = 0\), solve \(\frac{1}{64} t^2 (96 - t^2) = 0\) for \(t\):

\(t^2 = 96\), \(t = 4\sqrt{6}\).

Substitute \(t = 4\sqrt{6}\) into \(v = 16kt - \frac{1}{3} kt^3\):

\(v = \frac{3}{16} \times 16 \times 4\sqrt{6} - \frac{3}{16} \times \frac{1}{3} \times (4\sqrt{6})^3\)

\(v = 29.4 \text{ m s}^{-1}\)

(c) Maximum displacement occurs when \(v = 0\). Solve \(16kt - \frac{1}{3} kt^3 = 0\):

\(t^2 = 48\), \(t = 4\sqrt{3}\).

Substitute \(t = 4\sqrt{3}\) into \(s = \frac{1}{64} t^2 (96 - t^2)\):

\(s = \frac{1}{64} \times 48 \times (96 - 48)\)

\(s = 36 \text{ m}\)

(a) To find the distance \(AB\), we need to determine when the cyclist comes to rest at \(B\). The velocity \(v\) is the derivative of \(s\) with respect to \(t\):

\(v = \frac{ds}{dt} = 0.004(150t - 3t^2) = 0.6t - 0.012t^2\).

Set \(v = 0\) to find when the cyclist stops:

\(0.6t - 0.012t^2 = 0\)

\(t(0.6 - 0.012t) = 0\)

\(t = 0\) or \(t = 50\).

At \(t = 50\), substitute into \(s\):

\(s = 0.004(75 \times 50^2 - 50^3) = 0.004(187500 - 125000) = 0.004 \times 62500 = 250 \text{ m}\).

Thus, the distance \(AB = 250 \text{ m}\).

(b) The maximum velocity occurs when the acceleration is zero. Differentiate \(v\) to find acceleration \(a\):

\(a = \frac{dv}{dt} = 0.6 - 0.024t\).

Set \(a = 0\) to find the time of maximum velocity:

\(0.6 - 0.024t = 0\)

\(t = 25\).

Substitute \(t = 25\) into \(v\):

\(v = 0.004(150 \times 25 - 3 \times 25^2) = 0.004(3750 - 1875) = 0.004 \times 1875 = 7.5 \text{ ms}^{-1}\).

Thus, the maximum velocity is 7.5 \text{ ms}^{-1}.