(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}.

(a) To find the value of \(t\) at point B, we need to determine when the velocity \(v = 0\). The velocity is the integral of acceleration:

\(v = \int (6t^{\frac{1}{2}} - 2t) \, dt\)

\(v = 4t^{\frac{3}{2}} - t^2 + c\)

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

At point B, \(v = 0\):

\(4t^{\frac{3}{2}} - t^2 = 0\)

\(t^{\frac{1}{2}}(4t^{\frac{1}{2}} - t) = 0\)

\(t^{\frac{1}{2}} = 0 \text{ or } 4t^{\frac{1}{2}} = t\)

\(t = 0 \text{ or } t = 16\)

Thus, \(t = 16\) at point B.

(b) To find the distance travelled from A to the point where the acceleration is zero again, set \(a = 0\):

\(6t^{\frac{1}{2}} - 2t = 0\)

\(t^{\frac{1}{2}}(6 - 2t^{\frac{1}{2}}) = 0\)

\(t^{\frac{1}{2}} = 0 \text{ or } 6 = 2t^{\frac{1}{2}}\)

\(t = 0 \text{ or } t = 9\)

Integrate the velocity to find the distance \(s\):

\(s = \int (4t^{\frac{3}{2}} - t^2) \, dt\)

\(s = \left[ \frac{8}{5}t^{\frac{5}{2}} - \frac{1}{3}t^3 \right]\)

Evaluate from \(t = 0\) to \(t = 9\):

\(s = \frac{8}{5}(9)^{\frac{5}{2}} - \frac{1}{3}(9)^3\)

\(s = 145.8 \text{ m}\)

(a) To find the displacement, integrate the velocity function:

\(s = \int (t^2 - 8t^{3/2} + 10t) \, dt\)

\(s = \left[ \frac{t^3}{3} - \frac{16}{5}t^{5/2} + 5t^2 \right] + C\)

Using the limits from \(t = 0\) to \(t = 1\),

\(s = \left( \frac{1^3}{3} - \frac{16}{5} \times 1^{5/2} + 5 \times 1^2 \right) - \left( \frac{0^3}{3} - \frac{16}{5} \times 0^{5/2} + 5 \times 0^2 \right)\)

\(s = \frac{1}{3} - \frac{16}{5} + 5 = \frac{32}{15} \approx 2.13 \text{ m}\)

(b) To find the minimum velocity, differentiate the velocity function:

\(a = \frac{dv}{dt} = 2t - 12t^{1/2} + 10\)

Set \(a = 0\):

\(2t - 12t^{1/2} + 10 = 0\)

Solving this quadratic equation in terms of \(\sqrt{t}\),

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

\(t = 1\) or \(t = 25\)

Check the nature of the stationary point by differentiating \(a\):

\(\frac{da}{dt} = 2 - 6t^{-1/2}\)

At \(t = 25\), \(\frac{da}{dt} = 0.8\), indicating a minimum.

Evaluate \(v\) at \(t = 25\):

\(v = 25^2 - 8 \times 25^{3/2} + 10 \times 25 = -125 \text{ m s}^{-1}\)

\((a) To find when P is at instantaneous rest, set v = 0:\)

\(4t^2 - 20t + 21 = 0\)

Factorize the quadratic:

\((2t - 3)(2t - 7) = 0\)

\(Thus, t = 1.5 and t = 3.5.\)

(b) The acceleration a is given by the derivative of v:

\(a = \frac{dv}{dt} = 8t - 20\)

\(At t = 0, a = 8(0) - 20 = -20.\)

(c) To find the minimum velocity, set the derivative of v to zero:

\(8t - 20 = 0\)

Solve for t:

\(t = 2.5\)

Substitute back into v:

\(v = 4(2.5)^2 - 20(2.5) + 21 = -4\)

\(Thus, v_{min} = -4 \text{ ms}^{-1}.\)

(d) Integrate v to find the distance:

\(s = \int (4t^2 - 20t + 21) \, dt = \frac{4}{3}t^3 - 10t^2 + 21t + c\)

\(Evaluate from t = 1.5 to t = 3.5:\)

\(s = \left[ \frac{4}{3}(3.5)^3 - 10(3.5)^2 + 21(3.5) \right] - \left[ \frac{4}{3}(1.5)^3 - 10(1.5)^2 + 21(1.5) \right]\)

Calculate the distance:

\(s = \frac{49}{6} - \frac{27}{2} = \frac{16}{3} = 5.33 \text{ m}\)

(a) To find the velocity, integrate the acceleration:

\(v = \int 0.1t^{3/2} \, dt = 0.04t^{5/2} + 1.72\)

Set \(v = 3\) and solve for \(t\):

\(0.04t^{5/2} + 1.72 = 3\)

\(0.04t^{5/2} = 1.28\)

\(t^{5/2} = 32\)

\(t = 4\)

(b) To find the displacement, integrate the velocity:

\(s = \int (0.04t^{5/2} + 1.72) \, dt\)

\(s = \left[ \frac{2}{175}t^{7/2} + 1.72t \right]\)

Evaluate from \(t = 0\) to \(t = 2\):

\(s = \left[ \frac{2}{175}(2)^{7/2} + 1.72(2) \right]\)

\(s = 3.57 \text{ m}\)

Given the acceleration function: \(a = 6t - 18\).

To find the velocity \(v\), integrate the acceleration:

\(v = \int (6t - 18) \, dt = 3t^2 - 18t + C\).

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

Thus, \(v = 3t^2 - 18t\).

To find when the particle comes to rest, set \(v = 0\):

\(3t^2 - 18t = 0\).

Factor the equation: \(3t(t - 6) = 0\).

So, \(t = 0\) or \(t = 6\). The particle comes to rest at \(t = 6\) (ignoring \(t = 0\) as it is the starting point).

To find the distance \(s\), integrate the velocity:

\(s = \int (3t^2 - 18t) \, dt = t^3 - 9t^2 + C\).

Since \(s = 0\) when \(t = 0\), \(C = 0\).

Thus, \(s = t^3 - 9t^2\).

Substitute \(t = 6\) to find the distance:

\(s = 6^3 - 9 \times 6^2 = 216 - 324 = -108\).

Since distance must be positive, \(s = 108\) m.

(a) To find the velocity when the acceleration is zero, first find the acceleration by differentiating the velocity function:

\(a = \frac{dv}{dt} = \frac{d}{dt}(4.5 + 4t - 0.5t^2) = 4 - t\).

Set the acceleration to zero:

\(4 - t = 0\).

Solving for \(t\), we get \(t = 4\).

Substitute \(t = 4\) into the velocity equation:

\(v = 4.5 + 4(4) - 0.5(4)^2 = 12.5\) m/s.

(b) The particle comes to rest when \(v = 0\):

\(4.5 + 4t - 0.5t^2 = 0\).

Solving the quadratic equation:

\(0.5t^2 - 4t - 4.5 = 0\).

Using the quadratic formula, \(t = 9\) (reject \(t = -1\)).

To find the distance \(PQ\), integrate the velocity function from \(t = 0\) to \(t = 9\):

\(\int (4.5 + 4t - 0.5t^2) \, dt = 4.5t + 2t^2 - \frac{1}{6}t^3 \bigg|_0^9\).

Calculate the definite integral:

\([4.5(9) + 2(9)^2 - \frac{1}{6}(9)^3] - [4.5(0) + 2(0)^2 - \frac{1}{6}(0)^3] = 81\) m.

First, find the value of \(k\) by ensuring continuity of velocity at \(t = 8\):

\(30.6 - 0.9 \times 8 = \frac{1600}{8^2} + 8k\)

\(23.4 = 25 + 8k\)

\(k = -0.2\)

Next, find the time \(t\) when the particle comes to rest using \(v = \frac{1600}{t^2} + kt = 0\):

\(\frac{1600}{t^2} - 0.2t = 0\)

\(1600 = 0.2t^3\)

\(t^3 = 8000\)

\(t = 20\)

Now, integrate the velocity functions over the given intervals to find the distance:

For \(0 \leq t \leq 2\):

\(s = \int 7.2t^2 \, dt = \frac{7.2}{3}t^3 \bigg|_0^2 = 19.2 \text{ m}\)

For \(2 \leq t \leq 8\):

\(s = \int (30.6 - 0.9t) \, dt = (30.6t - \frac{0.9}{2}t^2) \bigg|_2^8 = 156.6 - 59.4 = 97.2 \text{ m}\)

For \(8 \leq t \leq 20\):

\(s = \int \left( \frac{1600}{t^2} - 0.2t \right) \, dt = \left( -\frac{1600}{t} - \frac{0.2}{2}t^2 \right) \bigg|_8^{20} = -120 - (-206.4) = 86.4 \text{ m}\)

Total distance \(OX = 19.2 + 97.2 + 86.4 = 262.2 \text{ m}\)

(a) To find the displacement \(s\), integrate the velocity function:

\(s = \int k(t^2 - 10t + 21) \, dt = k \left( \frac{1}{3} t^3 + 5t^2 + 21t \right) + C\)

Using the given displacements:

\(2.85 = k \left( \frac{1}{3} \times 3^3 - 5 \times 3^2 + 21 \times 3 \right) + C\)

\(2.4 = k \left( \frac{1}{3} \times 6^3 - 5 \times 6^2 + 21 \times 6 \right) + C\)

Solving these equations gives \(k = 0.05\) and \(C = 1.5\).

Thus, the displacement is:

\(s = 0.05 \left( \frac{1}{3} t^3 - 5t^2 + 21t \right) + 1.5\)

(b) To find when the velocity is a minimum, differentiate \(v\):

\(a = \frac{dv}{dt} = 0.05(2t - 10)\)

Set \(a = 0\) to find \(t = 5\).

Substitute \(t = 5\) into the displacement equation:

\(s = 0.05 \left( \frac{1}{3} \times 5^3 - 5 \times 5^2 + 21 \times 5 \right) + 1.5 = 2.58 \text{ m}\)

(a) To find when the particle is at rest, set the velocity \(v = \frac{ds}{dt}\) to zero. For \(0 \leq t \leq 6\), \(s = t^2 - 3t + 2\).

Differentiate: \(v = \frac{ds}{dt} = 2t - 3\).

Set \(v = 0\): \(2t - 3 = 0 \Rightarrow t = 1.5\).

(b) At \(t = 6\), find the velocity using \(s = \frac{24}{t} - \frac{t^2}{4} + 25\) for \(t \geq 6\).

Differentiate: \(v = -\frac{24}{t^2} - 0.5t\).

At \(t = 6\), \(v = -\frac{24}{6^2} - 0.5 \times 6 = 9 \text{ m/s}\).

Velocity when leaves: \(v = -3.67 \text{ m/s}\).

(c) Calculate displacement at key points:

• At \(t = 0\), \(s = 2\).
• At \(t = 1.5\), \(s = -0.25\).
• At \(t = 6\), \(s = 20\).
• At \(t = 10\), \(s = 2.4\).

Total distance = \(2 + 0.25 + 0.25 + 20 + (20 - 2.4) = 40.1 \text{ m}\).

(i) Given \(v = 0.04t^3 + ct^2 + kt\), at \(t = 5\), \(v = 10\).

\(10 = 0.04(5)^3 + c(5)^2 + 5k\)

\(10 = 0.04 \times 125 + 25c + 5k\)

\(10 = 5 + 25c + 5k\)

\(5c + k = 1\)

For distance, \(s = \int v \, dt = \frac{0.04}{4}t^4 + \frac{c}{3}t^3 + \frac{k}{2}t^2 + C\)

At \(t = 0, s = 0\), so \(C = 0\).

At \(t = 5, s = 25\):

\(25 = 0.01(5)^4 + \frac{125}{3}c + \frac{25}{2}k\)

\(25 = 0.01 \times 625 + \frac{125}{3}c + \frac{25}{2}k\)

\(25 = 6.25 + \frac{125}{3}c + \frac{25}{2}k\)

\(\frac{125}{3}c + \frac{25}{2}k = 18.75\)

Solving \(5c + k = 1\) and \(\frac{125}{3}c + \frac{25}{2}k = 18.75\) gives \(c = -0.3\) and \(k = 2.5\).

(ii) Acceleration \(a = \frac{dv}{dt} = 0.12t^2 - 0.6t + 2.5\).

To find minimum acceleration, set \(\frac{da}{dt} = 0\):

\(a' = 0.24t - 0.6 = 0\)

\(t = 2.5\)

Completing the square for \(a = 0.12(t^2 - 5t + ...) = 0.12((t - 2.5)^2 + ...)\) confirms minimum at \(t = 2.5\).

First, find the velocity \(v\) by differentiating the displacement \(s\) with respect to time \(t\):

\(v = \frac{ds}{dt} = 3t^2 - 12t + 4\).

Next, find the acceleration \(a\) by differentiating the velocity \(v\) with respect to time \(t\):

\(a = \frac{dv}{dt} = 6t - 12\).

Set the acceleration to zero to find the time \(t\) when the acceleration is zero:

\(6t - 12 = 0\)

\(t = 2\).

Substitute \(t = 2\) into the velocity equation to find the velocity at this time:

\(v = 3(2)^2 - 12(2) + 4 = 12 - 24 + 4 = -8 \text{ ms}^{-1}\).

(i) To find the expression for \(s\), we start by integrating the acceleration to find the velocity:

\(v = \int (6t - 12) \, dt = 3t^2 - 12t + C\)

Next, integrate the velocity to find the displacement:

\(s = \int (3t^2 - 12t + C) \, dt = t^3 - 6t^2 + Ct + D\)

Using the conditions \(s = 5\) when \(t = 1\) and \(s = 1\) when \(t = 3\), we set up the equations:

\(5 = 1^3 - 6(1)^2 + C(1) + D\)

\(1 = 3^3 - 6(3)^2 + 3C + D\)

Solving these gives \(C = 9\) and \(D = 1\). Thus, \(s = t^3 - 6t^2 + 9t + 1\).

(ii) The particle is at instantaneous rest when \(v = 0\):

\(3t^2 - 12t + 9 = 0\)

\(3(t^2 - 4t + 3) = 0\)

\((t - 1)(t - 3) = 0\)

Thus, \(t = 1\) or \(t = 3\).

(iii) To find the total distance travelled, evaluate \(s\) at key points:

For \(0 < t < 1\), \(s = (1 - 6 + 9 + 1) - 1 = 4\)

For \(1 < t < 3\), \(s = (27 - 54 + 27 + 1) - 5 = 4\)

For \(3 < t \leq 4\), \(s = (64 - 96 + 36 + 1) - 1 = 4\)

Total distance is \(4 + 4 + 4 = 12\) m.

(i) To find the minimum velocity, differentiate \(v = t^2 - 8t + 12\) to find the acceleration \(a\):

\(a = \frac{dv}{dt} = 2t - 8\)

Set \(a = 0\) to find the critical points:

\(2t - 8 = 0\)

\(t = 4\)

Substitute \(t = 4\) back into the velocity equation:

\(v = 4^2 - 8 \times 4 + 12 = 16 - 32 + 12 = -4 \text{ ms}^{-1}\)

Thus, the minimum velocity is \(-4 \text{ ms}^{-1}\).

(ii) To find the total distance, first find when \(v = 0\):

\(v = t^2 - 8t + 12 = 0\)

\((t - 2)(t - 6) = 0\)

\(t = 2\) or \(t = 6\)

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

\(s = \int (t^2 - 8t + 12) \, dt = \frac{1}{3}t^3 - 4t^2 + 12t + C\)

Calculate the displacement for each interval:

For \(0 \leq t \leq 2\):

\(s_1 = \left[ \frac{1}{3}t^3 - 4t^2 + 12t \right]_0^2 = \frac{8}{3} - 16 + 24 = \frac{32}{3}\)

For \(2 \leq t \leq 6\):

\(s_2 = \left[ \frac{1}{3}t^3 - 4t^2 + 12t \right]_2^6 = \left( \frac{216}{3} - 144 + 72 \right) - \left( \frac{8}{3} - 16 + 24 \right) = -\frac{32}{3}\)

For \(6 \leq t \leq 8\):

\(s_3 = \left[ \frac{1}{3}t^3 - 4t^2 + 12t \right]_6^8 = \left( \frac{512}{3} - 4 \times 8^2 + 12 \times 8 \right) - \left( \frac{216}{3} - 144 + 72 \right) = \frac{32}{3}\)

Total distance = \(|s_1| + |s_2| + |s_3| = \frac{32}{3} + \frac{32}{3} + \frac{32}{3} = 32 \text{ m}\).

(i) To find when the particle comes to instantaneous rest, we need to find when the velocity v is zero. The velocity is the integral of acceleration:

\(v = \int (0.4t^3 - 4.8t^{1/2}) \, dt = 0.1t^4 - 3.2t^{3/2} + c\)

\(Since the particle starts from rest, v = 0 when t = 0, so c = 0. Thus,\)

\(v = 0.1t^4 - 3.2t^{3/2}\)

\(Setting v = 0 gives:\)

\(0.1t^4 = 3.2t^{3/2}\)

Solving for t, we get:

\(t^{5/2} = 32\)

\(t = 4\)

\(Substitute t = 4 into the acceleration equation:\)

\(a = 0.4(4)^3 - 4.8(4)^{1/2} = 16\)

Thus, the acceleration is 16 m s^-2.

(ii) To find the displacement, integrate the velocity:

\(s = \int (0.1t^4 - 3.2t^{3/2}) \, dt = \left[0.02t^5 - 1.28t^{5/2}\right]_0^5\)

Calculate the displacement:

\(s = (0.02(5)^5 - 1.28(5)^{5/2}) - (0.02(0)^5 - 1.28(0)^{5/2})\)

\(s = -9.05\)

Thus, the displacement is -9.05 m.

(i) To find the maximum velocity, set the acceleration to zero:

\(25 - t^2 = 0\)

\(t^2 = 25\)

\(t = 5\)

Integrate the acceleration to find the velocity:

\(v = \int (25 - t^2) \, dt = 25t - \frac{1}{3}t^3 + C\)

Since the particle is initially at rest, \(v = 0\) when \(t = 0\), so \(C = 0\).

Substitute \(t = 5\) to find the maximum velocity:

\(v = 25 \times 5 - \frac{1}{3} \times 5^3 = 125 - \frac{125}{3} = \frac{250}{3} = 83\frac{1}{3} \text{ m s}^{-1}\)

(ii) To find the total distance travelled, integrate the velocity function from \(t = 0\) to \(t = 5\):

\(s = \int_0^5 (25t - \frac{1}{3}t^3) \, dt = \left[ \frac{25}{2}t^2 - \frac{1}{12}t^4 \right]_0^5\)

\(s = \left( \frac{25}{2} \times 25 - \frac{1}{12} \times 625 \right) = 260 \text{ m}\)

(iii) For \(t > 9\), the acceleration is given by \(a = -3t^{-1/2}\). Integrate to find the velocity change from \(t = 9\) to \(t = 25\):

At \(t = 9\), \(v = 25 \times 9 - \frac{1}{3} \times 9^3 = -18 \text{ m s}^{-1}\).

Integrate the acceleration:

\(\Delta v = \int_9^{25} -3t^{-1/2} \, dt = \left[ -6t^{1/2} \right]_9^{25}\)

\(\Delta v = -6 \times 5 + 6 \times 3 = -30 \text{ m s}^{-1}\)

Thus, the velocity at \(t = 25\) is \(-18 - 12 = -30 \text{ m s}^{-1}\).

(i) To find the time T when the particle reaches its maximum velocity, set the acceleration \(a = 0\):

\(1.2T^{1/2} - 0.6T = 0\)

Solving for \(T\):

\(1.2T^{1/2} = 0.6T\)

\(T^{1/2} = 0.5T\)

\(T^{1/2} = 2\)

\(T = 4\)

(ii) To find the velocity when the acceleration is maximum, first differentiate \(a\) with respect to \(t\):

\(\frac{da}{dt} = 0.6t^{-1/2} - 0.6\)

Set \(\frac{da}{dt} = 0\) to find \(t\):

\(0.6t^{-1/2} = 0.6\)

\(t^{-1/2} = 1\)

\(t = 1\)

Integrate \(a\) to find the velocity \(v\):

\(v = \int (1.2t^{1/2} - 0.6t) \, dt\)

\(v = 0.8t^{3/2} - 0.3t^2 + C\)

Using the initial condition \(v(0) = 1\):

\(1 = 0.8(0)^{3/2} - 0.3(0)^2 + C\)

\(C = 1\)

Thus, \(v = 0.8t^{3/2} - 0.3t^2 + 1\)

Substitute \(t = 1\) to find the velocity:

\(v = 0.8(1)^{3/2} - 0.3(1)^2 + 1\)

\(v = 0.8 - 0.3 + 1\)

\(v = 1.5\) m s^-1

(i) To find the time when the particle is again at instantaneous rest, we need to find when the velocity \(v = 0\). The acceleration is given by \(a = \frac{dv}{dt} = 6 - 0.24t\). Integrating with respect to \(t\), we get:

\(v = \int (6 - 0.24t) \, dt = 6t - 0.12t^2 + c\)

At \(t = 20\), the particle is at rest, so \(v = 0\):

\(0 = 6 \times 20 - 0.12 \times 20^2 + c\)

\(0 = 120 - 48 + c\)

\(c = -72\)

Now, set \(v = 0\) again to find the next time of rest:

\(0 = 6t - 0.12t^2 - 72\)

\(0.12t^2 - 6t + 72 = 0\)

Solving this quadratic equation gives \(t = 30\).

(ii) To find the distance traveled, we integrate the velocity function:

\(s = \int (6t - 0.12t^2 - 72) \, dt = 3t^2 - 0.04t^3 - 72t + k\)

Calculate \(s(30) - s(20)\):

\(s(30) = 3(30)^2 - 0.04(30)^3 - 72(30)\)

\(s(30) = 2700 - 1080 - 2160 = -540\)

\(s(20) = 3(20)^2 - 0.04(20)^3 - 72(20)\)

\(s(20) = 1200 - 320 - 1440 = -560\)

Distance traveled = \(|-540 - (-560)| = 20\).

(a) The velocity is given by integrating the acceleration:

\(v = \int a \, dt = \int kt^{1/2} \, dt = \frac{2}{3}kt^{3/2} + c\)

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

At \(t = 9\), \(v = 1.8\):

\(1.8 = \frac{2}{3}k \times 9^{3/2}\)

\(1.8 = \frac{2}{3}k \times 27\)

\(k = \frac{1.8 \times 3}{2 \times 27} = 0.1\)

(b) For \(t > 9\), the velocity is given by:

\(v = 0.2(t - 9)^2 + 1.8\)

The distance travelled from \(t = 0\) to \(t = 9\) is:

\(s_1 = \int_0^9 \frac{2}{3}kt^{3/2} \, dt = \frac{2}{3}k \left[ \frac{2}{5}t^{5/2} \right]_0^9\)

\(s_1 = \frac{4}{15}k \times 243 = \frac{4}{15} \times 0.1 \times 243 = 6.48 \text{ m}\)

The distance travelled from \(t = 9\) to \(t = 18\) is:

\(s_2 = \int_9^{18} (0.2(t - 9)^2 + 1.8) \, dt\)

\(s_2 = \left[ \frac{0.2}{3}(t - 9)^3 + 1.8t \right]_9^{18}\)

\(s_2 = \left[ \frac{0.2}{3} \times 9^3 + 1.8 \times 18 \right] - \left[ 0 + 1.8 \times 9 \right]\)

\(s_2 = \frac{0.2}{3} \times 729 + 32.4 - 16.2 = 64.8 \text{ m}\)

Thus, \(s_1 = \frac{1}{10}s_2\).

(c) For \(t > 9\), the acceleration is:

\(a = \frac{dv}{dt} = \frac{d}{dt}(0.2(t - 9)^2 + 1.8) = 0.4(t - 9)\)

The greatest acceleration occurs at \(t = 10\):

\(a = 0.4(10 - 9) = 0.4 \text{ m s}^{-2}\)

(i) To find the velocity \(v\), differentiate the displacement \(s\) with respect to \(t\):

\(v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 4t^2 + 4t) = 3t^2 - 8t + 4\).

(ii) To find when the particle is at instantaneous rest, set \(v = 0\):

\(3t^2 - 8t + 4 = 0\).

Solving this quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -8\), \(c = 4\):

\(t = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times 4}}{2 \times 3} = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6}\).

This gives \(t = \frac{12}{6} = 2\) and \(t = \frac{4}{6} = \frac{2}{3}\).

(iii) To find the minimum velocity, differentiate \(v\) and set the derivative to zero:

\(\frac{dv}{dt} = 6t - 8\).

Setting \(\frac{dv}{dt} = 0\) gives \(6t - 8 = 0\), so \(t = \frac{8}{6} = \frac{4}{3}\).

Substitute \(t = \frac{4}{3}\) back into the expression for \(v\):

\(v = 3\left(\frac{4}{3}\right)^2 - 8\left(\frac{4}{3}\right) + 4 = 3 \times \frac{16}{9} - \frac{32}{3} + 4\).

\(v = \frac{48}{9} - \frac{96}{9} + \frac{36}{9} = \frac{48 - 96 + 36}{9} = \frac{-12}{9} = -\frac{4}{3}\).

(i) To find k, integrate the acceleration to get the velocity:

\(v = \int k(3t^2 - 12t + 2) \, dt = k \left( \frac{3t^3}{3} - \frac{12t^2}{2} + 2t \right) + C\)

\(v = k(t^3 - 6t^2 + 2t) + C\)

Given initial velocity \(v = 0.4\) m/s when \(t = 0\), \(C = 0.4\).

At \(t = 1\), \(v = 0.1\):

\(0.1 = k(1 - 6 + 2) + 0.4\)

\(0.1 = k(-3) + 0.4\)

\(-0.3 = -3k\)

\(k = 0.1\)

(ii) Integrate the velocity to find displacement:

\(s = \int (0.1(t^3 - 6t^2 + 2t) + 0.4) \, dt\)

\(s = 0.1 \left( \frac{t^4}{4} - \frac{6t^3}{3} + \frac{2t^2}{2} \right) + 0.4t + C\)

\(s = 0.025t^4 - 0.2t^3 + 0.1t^2 + 0.4t\)

Since the particle starts from the origin, \(C = 0\).

(iii) Verify at \(t = 2\):

\(s = 0.025(2)^4 - 0.2(2)^3 + 0.1(2)^2 + 0.4(2)\)

\(s = 0.4 - 1.6 + 0.4 + 0.8 = 0\)

The particle is at the origin at \(t = 2\).

(i) To find when the particle is at rest, set \(v = 0\):

\(-0.01t^3 + 0.22t^2 - 0.4t = 0\)

Factor out \(t\):

\(t(-0.01t^2 + 0.22t - 0.4) = 0\)

Solving \(-0.01(t^2 - 22t + 40) = 0\) gives:

\(t^2 - 22t + 40 = 0\)

\((t - 20)(t - 2) = 0\)

Thus, \(t = 2\) or \(t = 20\).

(ii) Acceleration \(a\) is the derivative of velocity:

\(a = \frac{dv}{dt} = -0.03t^2 + 0.44t - 0.4\)

To find when acceleration is greatest, differentiate \(a\) and set to zero:

\(\frac{da}{dt} = -0.06t + 0.44 = 0\)

\(t = \frac{0.44}{0.06} = 7.33\)

(iii) To find the distance, integrate velocity from \(t = 2\) to \(t = 20\):

\(s(t) = \int (-0.01t^3 + 0.22t^2 - 0.4t) \, dt\)

\(s(t) = -\frac{0.01}{4}t^4 + \frac{0.22}{3}t^3 - 0.2t^2 + C\)

Evaluate from \(t = 2\) to \(t = 20\):

\(s(20) - s(2) = \frac{2673}{25} = 106.92\)

\(Distance = 107 m (rounded to nearest whole number).\)

(i) The velocity for \(0 \leq t \leq 5\) is given by \(v = 1.5 + 0.4t\). The acceleration is the derivative of velocity with respect to time, \(a = \frac{dv}{dt} = 0.4\) m s^-2.

(ii) For \(t \geq 5\), the velocity is \(v = \frac{100}{t^2} - 0.1t\). The particle is at rest when \(v = 0\). Setting \(\frac{100}{t^2} - 0.1t = 0\), we solve for \(t\):

\(\frac{100}{t^2} = 0.1t\)

\(100 = 0.1t^3\)

\(t^3 = 1000\)

\(t = 10 \text{ s}\)

(iii) For \(0 \leq t \leq 5\), the velocity is \(v = 1.5 + 0.4t\). The distance is the area under the velocity-time graph, which can be calculated using the trapezium rule:

\(\text{Distance} = \frac{1}{2} (1.5 + 3.5) \times 5 = 12.5 \text{ m}\)

For \(t \geq 5\), integrate the velocity function:

\(s(t) = \int \left( \frac{100}{t^2} - 0.1t \right) dt\)

\(= -\frac{100}{t} - 0.05t^2 + C\)

Calculate \(s(10) - s(5)\):

\(s(10) - s(5) = \left(-\frac{100}{10} - 0.05 \times 10^2\right) - \left(-\frac{100}{5} - 0.05 \times 5^2\right)\)

\(= (-10 - 5) - (-20 - 1.25)\)

\(= -15 + 21.25 = 6.25 \text{ m}\)

\(Total distance = 12.5 + 6.25 = 18.75 m.\)

(i) The acceleration a is given by the derivative of velocity with respect to time: \(a = \frac{dv}{dt} = 3 \times 2 \times (2t - 5)^2\). Setting \(a = 54\), we have:

\(6(2t - 5)^2 = 54\)

\((2t - 5)^2 = 9\)

\(2t - 5 = \pm 3\)

Solving for \(t\), we get \(t = 1\) or \(t = 4\).

(ii) The displacement s is the integral of velocity with respect to time: \(s = \int v \, dt = \int (2t - 5)^3 \, dt\).

Using the substitution \(u = 2t - 5\), \(du = 2 \, dt\), we have:

\(s = \frac{1}{2} \int u^3 \, du = \frac{1}{2} \times \frac{u^4}{4} + C = \frac{(2t - 5)^4}{8} + C\).

Given \(s = 0\) at \(t = 0\), we find \(C\) by substituting \(t = 0\):

\(0 = \frac{(-5)^4}{8} + C\)

\(C = -\frac{625}{8}\)

Thus, the displacement is \(s = \frac{(2t-5)^4}{8} - \frac{625}{8}\).

(i) The velocity is given by v = qt + rt². We know that v = 4 when t = 1 and t = 2. This gives us the equations:

\(1. q + r = 4 (when t = 1)
2. 2q + 4r = 4 (when t = 2)\)

\(Solving these equations, we find q = 6 and r = -2.\)

The acceleration a is the derivative of v with respect to t:

\(a = \frac{dv}{dt} = q + 2rt = 6 - 4t\)

\(When t = 0.5,\)

\(a = 6 - 4(0.5) = 4 \text{ m s}^{-2}\)

\((ii) The particle is at instantaneous rest when v = 0:\)

\(v = 6t - 2t^2 = 0\)

\(Solving, we get t(6 - 2t) = 0, so t = 0 or t = 3.\)

\((iii) To find the distance from O when t = 0, we integrate the velocity:\)

\(s = \int (6t - 2t^2) \, dt = 3t^2 - \frac{2}{3}t^3 + C\)

\(Given that s = 0 when t = 3,\)

\(0 = 3(3)^2 - \frac{2}{3}(3)^3 + C\)

\(0 = 27 - 18 + C\)

\(C = -9\)

\(Thus, the distance from O when t = 0 is:\)

\(s = 3(0)^2 - \frac{2}{3}(0)^3 - 9 = -9\)

The distance is 9 m.

To solve this problem, we need to find the velocity and acceleration of the particle by differentiating the displacement function.

The displacement function is given by \(s = 2t^2 - \frac{80}{3}t^{3/2}\).

(i) First, find the velocity \(v\) by differentiating \(s\) with respect to \(t\):

\(v = \frac{ds}{dt} = \frac{d}{dt}(2t^2 - \frac{80}{3}t^{3/2}) = 4t - 40t^{1/2}\).

Next, find the acceleration \(a\) by differentiating \(v\) with respect to \(t\):

\(a = \frac{dv}{dt} = \frac{d}{dt}(4t - 40t^{1/2}) = 4 - 20t^{-1/2}\).

Set the acceleration to zero to find the time:

\(4 - 20t^{-1/2} = 0\)

\(4 = 20t^{-1/2}\)

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

\(t^{1/2} = 5\)

\(t = 25\) s

(ii) Substitute \(t = 25\) into the displacement and velocity equations:

Displacement: \(s = 2(25)^2 - \frac{80}{3}(25)^{3/2}\)

\(s = 2(625) - \frac{80}{3}(125)\)

\(s = 1250 - \frac{10000}{3}\)

\(s = 1250 - 3333.33\)

\(s = -2083.3\) m

Velocity: \(v = 4(25) - 40(25)^{1/2}\)

\(v = 100 - 40(5)\)

\(v = 100 - 200\)

\(v = -100\) m/s

(i) To find the maximum acceleration, differentiate \(a = 15t - 3t^2\) with respect to \(t\):

\(\frac{da}{dt} = 15 - 6t\)

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

\(15 - 6t = 0\)

\(t = 2.5\) seconds.

Substitute \(t = 2.5\) into \(a = 15t - 3t^2\):

\(a = 15(2.5) - 3(2.5)^2 = 18.75\) m s\(^{-2}\).

(ii) To find the distance, first find the velocity by integrating the acceleration:

\(v(t) = \int (15t - 3t^2) \, dt = 7.5t^2 - t^3 + c\)

Since the car starts from rest, \(v(0) = 0\), so \(c = 0\).

Now integrate the velocity to find the distance:

\(s(t) = \int (7.5t^2 - t^3) \, dt = 2.5t^3 - 0.25t^4 + d\)

Since \(s(0) = 0\), \(d = 0\).

At \(t = 5\),

\(s(5) = 2.5(5)^3 - 0.25(5)^4 = 156.25\) m.

(iii) The car comes to rest when \(v(k) = 0\). First, find \(v(5)\):

\(v(5) = 7.5(5)^2 - (5)^3 = 62.5\) m s\(^{-1}\).

For \(5 < t \leq k\), integrate \(a = -\frac{625}{t^2}\) to find \(v(t)\):

\(v(t) = \int -\frac{625}{t^2} \, dt = \frac{625}{t} + c\)

Using \(v(5) = 62.5\),

\(62.5 = \frac{625}{5} + c\)

\(c = -62.5\).

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

\(\frac{625}{k} - 62.5 = 0\)

\(\frac{625}{k} = 62.5\)

\(k = 10\).

\((i) The particle is decelerating when the acceleration is negative. Given a = 6t - 2, set 6t - 2 < 0:\)

\(6t - 2 < 0\)

\(6t < 2\)

\(t < \frac{1}{3}\)

\(Thus, the particle is decelerating for 0 < t < \frac{1}{3}.\)

(ii) To find s in terms of t, first find the velocity v by integrating the acceleration:

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

Next, integrate the velocity to find the displacement s:

\(s = \int (3t^2 - 2t + c) \, dt = t^3 - t^2 + ct + d\)

\(Using the conditions t = 1, s = 7 and t = 3, s = 29, form the equations:\)

\(c + d = 7\)

\(3c + d = 11\)

\(Solving these simultaneous equations gives c = 2 and d = 5.\)

\(Thus, s = t^3 - t^2 + 2t + 5.\)

\((iii) To find the time when the velocity is 10 m s^-1, set v = 10:\)

\(3t^2 - 2t + 2 = 10\)

\(3t^2 - 2t - 8 = 0\)

\(Solving this quadratic equation gives t = 2.\)

(i) To find when P is at instantaneous rest, set the velocity v to zero:

\(4t^2 - 8t + 3 = 0\)

Factor the quadratic equation:

\((2t - 3)(2t - 1) = 0\)

Solving gives \(t = 0.5\) and \(t = 1.5\).

(ii) To find the distance travelled, integrate the velocity function:

\(s = -\int (4t^2 - 8t + 3) \, dt\)

Integrate to find the displacement:

\(s = -\left[ \frac{4}{3}t^3 - 4t^2 + 3t \right]_{0.5}^{1.5}\)

Evaluate the definite integral:

\(s = -\left[ \left( \frac{4}{3}(1.5)^3 - 4(1.5)^2 + 3(1.5) \right) - \left( \frac{4}{3}(0.5)^3 - 4(0.5)^2 + 3(0.5) \right) \right]\)

\(s = -\left[ 0 - \left( -\frac{2}{3} \right) \right] = \frac{2}{3}\) m

(a) We have the equations from the given conditions:

\(4b + 4^{\frac{3}{2}}c = 8\)

\(9b + 9^{\frac{3}{2}}c = 13.5\)

Solving these simultaneously, we find \(b = 3\) and \(c = -0.5\).

(b) The acceleration \(a\) is given by \(a = \frac{dv}{dt} = b + \frac{3}{2}ct^{\frac{1}{2}}\).

Substituting \(b = 3\), \(c = -0.5\), and \(t = 1\), we get:

\(a = 3 + \frac{3}{2}(-0.5)(1)^{\frac{1}{2}} = 3 - 0.75 = 2.25 \text{ m/s}^2\)

(c) For instantaneous rest, set \(v = 0\):

\(3t - 0.5t^{\frac{3}{2}} = 0\)

Solving gives \(t = 36\).

To find the distance, integrate \(v\) to get \(s\):

\(s = \int (3t - 0.5t^{\frac{3}{2}}) \, dt = \frac{3}{2}t^2 - \frac{1}{5}t^{\frac{5}{2}} + C\)

Using limits from 0 to 36, \(s = 388.8 \text{ m}\).

(d) For the speed when returning to \(O\), solve:

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

\(t = 56.25\).

Substitute back into \(v\):

\(v = 3(56.25) - 0.5(56.25)^{\frac{3}{2}} = 42.2 \text{ m/s}\)

(i) The acceleration \(a\) is the derivative of the velocity \(v\). So, \(a = \frac{dv}{dt} = 12t - 30\). For the acceleration to be negative, \(12t - 30 < 0\). Solving this inequality gives \(t < 2.5\).

(ii) The particle is at instantaneous rest when \(v = 0\). Solving \(6t^2 - 30t + 24 = 0\) gives \(t = 1\) and \(t = 4\). The distance \(s\) is the integral of \(v\), so \(s = \int (6t^2 - 30t + 24) \, dt = 2t^3 - 15t^2 + 24t\). Evaluating from \(t = 1\) to \(t = 4\), \(s = [2t^3 - 15t^2 + 24t]_1^4\), which gives a distance of 27 m.

(iii) The particle passes through \(O\) when \(s = 0\). Solving \(2t^3 - 15t^2 + 24t = 0\) gives \(t = 0\), \(t = 2.31\), and \(t = 5.19\). The positive values are \(t = 2.31\) and \(t = 5.19\).

(i) To find when P returns to O, set \(x = 0\):

\(0.08t^2 - 0.0002t^3 = 0\)

\(t^2(0.08 - 0.0002t) = 0\)

\(t = 0\) or \(0.08 - 0.0002t = 0\)

\(0.0002t = 0.08\)

\(t = \frac{0.08}{0.0002} = 400\) s

To find the speed at O on return, differentiate \(x\) to get velocity \(v\):

\(v = \frac{dx}{dt} = 0.16t - 0.0006t^2\)

At \(t = 400\):

\(v = 0.16 \times 400 - 0.0006 \times 400^2\)

\(v = 64 - 96 = -32\) m/s

Speed is 32 m/s.

(ii)(a) To find the total distance travelled, find the time to the furthest point:

\(v = 0.16t - 0.0006t^2 = 0\)

\(t(0.16 - 0.0006t) = 0\)

\(t = 0\) or \(t = \frac{0.16}{0.0006} \approx 266.67\) s

Distance to furthest point:

\(x = 0.08(266.67)^2 - 0.0002(266.67)^3\)

\(x \approx 1895\) m

Total distance = \(2 \times 1895 = 3790\) m

(ii)(b) Average speed = \(\frac{\text{total distance}}{\text{total time}} = \frac{3790}{400} = 9.48\) m/s

(i) To find the acceleration, differentiate the velocity function \(v(t) = 0.00004t^3 - 0.006t^2 + 0.288t\) with respect to \(t\):

\(a(t) = \frac{dv}{dt} = 0.00012t^2 - 0.012t + 0.288\).

Set \(a(t) = 0\) to find when the acceleration is zero:

\(0.00012t^2 - 0.012t + 0.288 = 0\).

Factor the quadratic equation:

\(0.00012(t^2 - 100t + 2400) = 0.00012(t - 40)(t - 60) = 0\).

Thus, \(t = 40\) and \(t = 60\).

(ii) To find the displacement, integrate the velocity function from \(t = 0\) to \(t = 100\):

\(\int_0^{100} (0.00004t^3 - 0.006t^2 + 0.288t) \, dt\).

Calculate the integral:

\(\left[ 0.00001t^4 - 0.002t^3 + 0.144t^2 \right]_0^{100}\).

Evaluate at the limits:

\(0.00001(100)^4 - 0.002(100)^3 + 0.144(100)^2 = 440\).

Therefore, the displacement is 440 m.

(i) To find the displacement \(s\), integrate the velocity function:

\(s = \int (0.6t - 0.03t^2) \, dt = 0.3t^2 - 0.01t^3 + C\).

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

Calculate \(s(5)\):

\(s(5) = 0.3 \times 5^2 - 0.01 \times 5^3 = 6.25\).

To find the acceleration, differentiate the velocity function:

\(a = \frac{dv}{dt} = 0.6 - 0.06t\).

Calculate \(a(5)\):

\(a(5) = 0.6 - 0.06 \times 5 = 0.3 \text{ m s}^{-2}\).

(ii) Maximum velocity occurs when \(a = 0\):

\(0.6 - 0.06t = 0 \Rightarrow t = 10\).

Maximum velocity is \(v(10) = 3 \text{ m s}^{-1}\).

Half of maximum velocity is \(1.5 \text{ m s}^{-1}\).

Set \(0.6t - 0.03t^2 = 1.5\):

\(0.03t^2 - 0.6t + 1.5 = 0\).

Solving the quadratic equation:

\(t^2 - 20t + 50 = 0\).

Using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), find:

\(t = 2.93 \text{ s}\) and \(t = 17.07 \text{ s}\).

To find the displacement, we first integrate the acceleration to find the velocity. Given:

\(a(t) = 0.075t^2 - 1.5t + 5\)

Integrate \(a(t)\) to find \(v(t)\):

\(v(t) = \int (0.075t^2 - 1.5t + 5) \, dt = 0.025t^3 - 0.75t^2 + 5t + C\)

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

Thus, \(v(t) = 0.025t^3 - 0.75t^2 + 5t\).

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

\(s(t) = \int (0.025t^3 - 0.75t^2 + 5t) \, dt = 0.00625t^4 - 0.25t^3 + 2.5t^2 + C\)

Since the displacement is zero at \(t = 0\), \(C = 0\).

Thus, \(s(t) = 0.00625t^4 - 0.25t^3 + 2.5t^2\).

To find when \(s(t) = 0\) again, solve:

\(0.00625t^4 - 0.25t^3 + 2.5t^2 = 0\)

Factor out \(t^2\):

\(t^2(0.00625t^2 - 0.25t + 2.5) = 0\)

\(t^2 = 0\) gives \(t = 0\), and solving \(0.00625t^2 - 0.25t + 2.5 = 0\) gives \(t = 20\).

Therefore, the time taken for P to return to the point O is 20 seconds.

(i) For the first 8 seconds, the velocity \(v(8) = 0.25 \times 8 = 2\) m/s. Using the given velocity equation:

\(v = -0.1t^2 + 2.4t - k\)

Substitute \(t = 8\):

\(2 = -0.1(8)^2 + 2.4(8) - k\)

\(2 = -6.4 + 19.2 - k\)

\(k = 10.8\)

(ii) To find the maximum velocity, differentiate \(v\) with respect to \(t\) and set the derivative to zero:

\(\frac{dv}{dt} = -0.2t + 2.4\)

Set \(\frac{dv}{dt} = 0\):

\(-0.2t + 2.4 = 0\)

\(t = 12\)

Substitute \(t = 12\) into the velocity equation:

\(v = -0.1(12)^2 + 2.4(12) - 10.8\)

\(v = -14.4 + 28.8 - 10.8\)

\(v = 3.6\) m/s

(iii) Displacement for the first 8 seconds:

\(s_1 = \frac{1}{2} \times 0.25 \times 8^2 = 8\) m

Displacement from \(t = 8\) to \(t = 18\):

\(s_2 = \int_{8}^{18} (-0.1t^2 + 2.4t - 10.8) \, dt\)

\(s_2 = \left[ -0.1 \frac{t^3}{3} + 1.2t^2 - 10.8t \right]_{8}^{18}\)

\(s_2 = [-0.1(18)^3/3 + 1.2(18)^2 - 10.8(18)] - [-0.1(8)^3/3 + 1.2(8)^2 - 10.8(8)]\)

\(s_2 = 26.7\) m

Total displacement \(s = s_1 + s_2 = 8 + 26.7 = 34.7\) m

(i) For \(t = 3\), substitute into the velocity equation: \(v = -0.2(3)^2 + 4(3) - 15 = -4.8\) m s^-1. The acceleration for the first 3 s is constant, using \(v = u + at\) with \(u = 0\), \(v = -4.8\), and \(t = 3\), gives \(a = \frac{-4.8}{3} = -1.6\) m s^-2.

(ii) To find the maximum velocity, set \(\frac{dv}{dt} = 0\). Differentiate \(v = -0.2t^2 + 4t - 15\) to get \(\frac{dv}{dt} = -0.4t + 4\). Solve \(-0.4t + 4 = 0\) to find \(t = 10\). Substitute \(t = 10\) into the velocity equation: \(v = -0.2(10)^2 + 4(10) - 15 = 5\) m s^-1.

(iii) (a) Distance from 0 to 3 s is \(\frac{1}{2} \times 3 \times 4.8 = 7.2\) m. From 3 to 5 s, integrate \(v = -0.2t^2 + 4t - 15\) from 3 to 5: \(\int_{3}^{5} (-0.2t^2 + 4t - 15) \, dt = 4.533\) m. Total distance = 7.2 + 4.533 = 11.733 m. Average speed = \(\frac{11.733}{5} = 2.35\) m s^-1.

(b) For the whole journey, integrate from 3 to 15: \(\int_{3}^{15} (-0.2t^2 + 4t - 15) \, dt = 45.066\) m. Total distance = 7.2 + 45.066 = 52.266 m. Average speed = \(\frac{52.266}{15} = 3.00\) m s^-1.

(i) For \(t > 8\), the velocity \(v(t) = \frac{1}{2} t^{2/3}\). The acceleration \(a(t)\) is the derivative of \(v(t)\) with respect to \(t\):

\(a(t) = \frac{d}{dt} \left( \frac{1}{2} t^{2/3} \right) = \frac{1}{3} t^{-1/3}\).

At \(t = 8\), the acceleration just before passing \(A\) is \(\frac{1}{4} \text{ m s}^{-2}\) and just after is \(\frac{1}{3} \times 8^{-1/3} = \frac{1}{6} \text{ m s}^{-2}\).

The decrease in acceleration is \(\frac{1}{4} - \frac{1}{6} = \frac{1}{12} \text{ m s}^{-2}\).

(ii) The distance moved from \(t = 0\) to \(t = 8\) is given by:

\(s_1 = \frac{1}{2} \times \frac{1}{4} \times 8^2 = 8 \text{ m}\).

For \(t > 8\), the distance \(s_2\) is:

\(s_2 = \int_{8}^{27} \frac{1}{2} t^{2/3} \, dt = \left[ 0.3 t^{5/3} \right]_{8}^{27}\).

Calculating the definite integral:

\(s_2 = 0.3 \times 27^{5/3} - 0.3 \times 8^{5/3} = 71.3 - 8 = 63.3 \text{ m}\).

Total distance \(s = s_1 + s_2 = 8 + 63.3 = 71.3 \text{ m}\).

(i) To find \(k_1\), use the formula for distance: \(s = \int v \, dt\). For \(0 \leq t \leq 60\),

\(s = \int_0^{60} (k_1 t - 0.005t^2) \, dt = \left[ \frac{k_1 t^2}{2} - 0.005 \frac{t^3}{3} \right]_0^{60} = 540\)

\(k_1 \frac{60^2}{2} - 0.005 \frac{60^3}{3} = 540\)

\(1800k_1 - 72 = 540\)

\(1800k_1 = 612\)

\(k_1 = 0.5\)

For \(k_2\), equate \(v(60)\) from both expressions:

\(v(60) = 0.5 \times 60 - 0.005 \times 60^2 = \frac{k_2}{\sqrt{60}}\)

\(30 - 18 = \frac{k_2}{\sqrt{60}}\)

\(12 = \frac{k_2}{\sqrt{60}}\)

\(k_2 = 12\sqrt{60}\)

(ii) For \(t \geq 60\), the distance is:

\(s = 540 + \int_{60}^{t} \frac{12\sqrt{60}}{\sqrt{t}} \, dt\)

\(s = 540 + 12\sqrt{60} \left[ 2\sqrt{t} - 2\sqrt{60} \right]\)

\(s = 24\sqrt{60t} - 900\)

(iii) Set \(s(t) = 1260\):

\(24\sqrt{60t} - 900 = 1260\)

\(24\sqrt{60t} = 2160\)

\(\sqrt{60t} = 90\)

\(60t = 8100\)

\(t = 135\)

Find speed at \(t = 135\):

\(v = \frac{12\sqrt{60}}{\sqrt{135}} = 8 \text{ m s}^{-1}\)

(i) For 0 ≤ t ≤ 5, the velocity is given by v = t - 0.1t². The distance travelled, s_1, is the integral of velocity with respect to time:

\(s_1 = \int_0^5 (t - 0.1t^2) \, dt = \left[ \frac{t^2}{2} - \frac{0.1t^3}{3} \right]_0^5\)

\(s_1 = \left( \frac{25}{2} - \frac{0.1 \times 125}{3} \right) = 8.33 \text{ m}\)

(ii) For 5 ≤ t ≤ 45, the velocity is constant. Let the constant velocity be v_c. Since the particle starts from rest and returns to rest, the symmetry implies v_c = 2.5 m/s. The distance travelled, s_2, is:

\(s_2 = v_c \times (45 - 5) = 2.5 \times 40 = 100 \text{ m}\)

\(For 45 ≤ t ≤ 50, the velocity is given by v = 9t - 0.1t^2 - 200. The distance travelled, s_3, is:\)

\(s_3 = \int_{45}^{50} (9t - 0.1t^2 - 200) \, dt\)

\(s_3 = \left[ \frac{9t^2}{2} - \frac{0.1t^3}{3} - 200t \right]_{45}^{50}\)

\(s_3 = \left( \frac{9 \times 2500}{2} - \frac{0.1 \times 125000}{3} - 200 \times 50 \right) - \left( \frac{9 \times 2025}{2} - \frac{0.1 \times 91125}{3} - 200 \times 45 \right)\)

\(s_3 = 8.33 \text{ m}\)

The total distance from O to A is:

\(s = s_1 + s_2 + s_3 = 8.33 + 100 + 8.33 = 117 \text{ m}\)

The average speed is:

\(\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{117}{50} = 2.33 \text{ m/s}\)

To find when the particle is at rest, we need to find when the velocity \(v = \frac{ds}{dt}\) is zero.

Differentiate \(s = t^{\frac{5}{2}} - \frac{15}{4} t^{\frac{3}{2}} + 6\) with respect to \(t\):

\(v = \frac{5}{2} t^{\frac{3}{2}} - \frac{45}{8} t^{\frac{1}{2}}\).

Set \(v = 0\):

\(\frac{5}{2} t^{\frac{3}{2}} - \frac{45}{8} t^{\frac{1}{2}} = 0\).

Factor out \(t^{\frac{1}{2}}\):

\(t^{\frac{1}{2}} \left( \frac{5}{2} t - \frac{45}{8} \right) = 0\).

Since \(t^{\frac{1}{2}} = 0\) is not possible (as \(t = 0\) is the starting point), solve:

\(\frac{5}{2} t - \frac{45}{8} = 0\).

\(5t = \frac{45}{4}\).

\(t = \frac{45}{20} = 2.25\).

Substitute \(t = 2.25\) back into the displacement equation:

\(s = (2.25)^{\frac{5}{2}} - \frac{15}{4} (2.25)^{\frac{3}{2}} + 6\).

Calculate \(s\):

\(s = \frac{15}{16}\).

(i) The acceleration is given by \(a(t) = 0.6t\). Integrating to find velocity:

\(v(t) = \int 0.6t \, dt = 0.3t^2 + C\).

Since the particle starts from rest, \(v(0) = 0\), so \(C = 0\). Thus, \(v(t) = 0.3t^2\).

At \(t = 10\), \(v(10) = 0.3 \times 10^2 = 30\) m s⁻¹.

Integrating velocity to find displacement:

\(s(t) = \int 0.3t^2 \, dt = 0.1t^3 + C\).

Since \(s(0) = 0\), \(C = 0\). Thus, \(s(t) = 0.1t^3\).

At \(t = 10\), \(s(10) = 0.1 \times 10^3 = 100\) m.

(ii) After \(t = 10\), the acceleration is \(a(t) = -0.4t\). Integrating to find velocity:

\(v(t) = \int -0.4t \, dt = -0.2t^2 + C\).

Using \(v(10) = 30\), \(-0.2 \times 10^2 + C = 30\) gives \(C = 50\). Thus, \(v(t) = -0.2t^2 + 50\).

At point \(A\), \(v(t) = 0\):

\(-0.2t^2 + 50 = 0\) gives \(t = \sqrt{250}\).

Integrating velocity to find displacement:

\(s(t) = \int (-0.2t^2 + 50) \, dt = -\frac{t^3}{15} + 50t + C\).

Using \(s(10) = 100\), \(-\frac{10^3}{15} + 500 + C = 100\) gives \(C = -\frac{1000}{3}\).

Thus, \(s(t) = -\frac{t^3}{15} + 50t - \frac{1000}{3}\).

At \(t = \sqrt{250}\), \(s(\sqrt{250}) = -\frac{(\sqrt{250})^3}{15} + 50\sqrt{250} - \frac{1000}{3} = 194\) m.

(i) To find the time t when the aeroplane takes off, we need to find the velocity function by integrating the acceleration function. The acceleration is given by:

\(a(t) = 1.5 + 0.012t\)

Integrating with respect to t gives the velocity function:

\(v(t) = \int (1.5 + 0.012t) \, dt = 1.5t + 0.006t^2 + C\)

Since the aeroplane starts from rest, \(v(0) = 0\), so \(C = 0\).

Thus, the velocity function is:

\(v(t) = 1.5t + 0.006t^2\)

We set \(v(t) = 90\) to find the time when the aeroplane takes off:

\(1.5t + 0.006t^2 = 90\)

Solving the quadratic equation:

\(0.006t^2 + 1.5t - 90 = 0\)

\(t^2 + 250t - 15000 = 0\)

\((t - 50)(t + 300) = 0\)

Thus, \(t = 50\) (since time cannot be negative).

(ii) To find the distance travelled, integrate the velocity function:

\(s(t) = \int (1.5t + 0.006t^2) \, dt = 0.75t^2 + 0.002t^3 + C\)

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

Thus, the distance function is:

\(s(t) = 0.75t^2 + 0.002t^3\)

Substitute \(t = 50\) to find the distance:

\(s(50) = 0.75(50)^2 + 0.002(50)^3\)

\(= 0.75(2500) + 0.002(125000)\)

\(= 1875 + 250\)

\(= 2125\) meters.

(i) To find the acceleration, differentiate the velocity function:

\(a(t) = \frac{dv}{dt} = \frac{d}{dt}(0.2t + 0.006t^2) = 0.2 + 0.012t\).

The initial acceleration is \(a(0) = 0.2\).

We are given that \(a(t) = 2.5 \times a(0)\), so:

\(0.2 + 0.012t = 2.5 \times 0.2\).

Simplifying gives:

\(0.2 + 0.012t = 0.5\).

\(0.012t = 0.3\).

\(t = \frac{0.3}{0.012} = 25\).

(ii) To find the displacement, integrate the velocity function:

\(s(t) = \int (0.2t + 0.006t^2) \, dt = 0.1t^2 + 0.002t^3 + C\).

Assuming \(C = 0\) (since the particle starts from O), evaluate from \(t = 0\) to \(t = 25\):

\(s(25) = 0.1 \times 25^2 + 0.002 \times 25^3\).

\(s(25) = 0.1 \times 625 + 0.002 \times 15625\).

\(s(25) = 62.5 + 31.25 = 93.75\).

(i) To find \(k\), we need to determine when the velocity \(v(t) = k(60t^2 - t^3)\) is at its maximum. The derivative \(\frac{dv}{dt} = k(120t - 3t^2)\) is set to zero for maximum velocity:

\(120t - 3t^2 = 0\)

\(t(120 - 3t) = 0\)

\(t = 0 \text{ or } t = 40\)

At \(t = 40\), \(v(40) = k(60 \times 40^2 - 40^3) = 6.4\)

\(k(96000 - 64000) = 6.4\)

\(k \times 32000 = 6.4\)

\(k = \frac{6.4}{32000} = 0.0002\)

(ii) The particle comes to rest at \(t = 60\). Integrate \(v(t)\) to find \(s(t)\):

\(s(t) = \int k(60t^2 - t^3) \, dt = 0.0002 \left( 20t^3 - \frac{t^4}{4} \right) + C\)

Using \(t = 60\) and \(C = 0\):

\(OA = 0.0002 \left( 20 \times 60^3 - \frac{60^4}{4} \right)\)

\(OA = 0.0002 \times 1296000 = 216 \text{ m}\)

(iii) The acceleration at \(t = 60\) is given by:

\(\frac{dv}{dt} = 0.0002(120 \times 60 - 3 \times 60^2)\)

\(= 0.0002 \times 3600 = 0.72 \text{ m s}^{-2}\)

(iv) To find the speed when \(P\) passes through \(O\) again, solve \(s(t) = 0\) for non-zero \(t\):

\(20t^3 - \frac{t^4}{4} = 0\)

\(t^3(20 - \frac{t}{4}) = 0\)

\(t = 0 \text{ or } t = 80\)

At \(t = 80\), \(v(80) = 0.0002(60 \times 80^2 - 80^3)\)

\(= 0.0002(384000 - 512000)\)

\(= 0.0002 \times (-128000) = -25.6 \text{ m s}^{-1}\)

The speed is 25.6 m s^-1.

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) for the segment \(AB\):

\(55 = 5u + \frac{1}{2}a(5^2)\)

\(55 = 5u + 12.5a\) (Equation 1)

For the segment \(BC\), the final velocity at \(B\) is \(v_B = u + 5a\). Using the equation \(s = ut + \frac{1}{2}at^2\) again:

\(65 = 5v_B + \frac{1}{2}a(5^2)\)

\(65 = 5(u + 5a) + 12.5a\)

\(65 = 5u + 25a + 12.5a\)

\(65 = 5u + 37.5a\) (Equation 2)

Subtract Equation 1 from Equation 2:

\(65 - 55 = (5u + 37.5a) - (5u + 12.5a)\)

\(10 = 25a\)

\(a = 0.4 \text{ m s}^{-2}\)

Substitute \(a = 0.4\) into Equation 1:

\(55 = 5u + 12.5(0.4)\)

\(55 = 5u + 5\)

\(50 = 5u\)

\(u = 10 \text{ m s}^{-1}\)

(i) To find the distance OA, we integrate the velocity function to find the displacement:

\(s = \int v \, dt = \int 0.027(10t^2 - t^3) \, dt\)

\(s = 0.027 \left( \frac{10t^3}{3} - \frac{t^4}{4} \right) + C\)

Since the particle starts from O, we set the constant of integration C to zero. At point A, the particle comes to rest, so we solve for t when v = 0:

\(0.027(10t^2 - t^3) = 0\)

\(t^2(10 - t) = 0\)

\(Thus, t = 10 s (ignoring t = 0 as it corresponds to the starting point).\)

\(Substitute t = 10 into the displacement equation:\)

\(s = 0.027 \left( \frac{10(10)^3}{3} - \frac{(10)^4}{4} \right)\)

\(s = 0.027 \left( \frac{10000}{3} - 2500 \right)\)

\(s = 0.027 \left( 3333.33 - 2500 \right)\)

\(s = 0.027 \times 833.33 = 22.5 \text{ m}\)

(ii) To find the maximum velocity, we differentiate the velocity function and set the derivative to zero:

\(\frac{dv}{dt} = 0.027(20t - 3t^2)\)

Set \(\frac{dv}{dt} = 0\):

\(0.027(20t - 3t^2) = 0\)

\(20t - 3t^2 = 0\)

\(t(20 - 3t) = 0\)

Thus, t = 0 or t = \(\frac{20}{3}\) s.

Substitute t = \(\frac{20}{3}\) into the velocity equation:

\(v_{\text{max}} = 0.027 \left( 10 \left( \frac{20}{3} \right)^2 - \left( \frac{20}{3} \right)^3 \right)\)

\(v_{\text{max}} = 0.027 \left( \frac{4000}{9} - \frac{8000}{27} \right)\)

\(v_{\text{max}} = 0.027 \times \frac{4000 - 2962.96}{9}\)

\(v_{\text{max}} = 0.027 \times 148.15 = 4 \text{ m s}^{-1}\)

(i) To find the velocity \(v(t)\), integrate the acceleration function \(a(t) = 2t^{2/3}\):

\(v(t) = \int 2t^{2/3} \, dt = 1.2t^{5/3} + C\)

Given that the initial velocity \(v(0) = 2\), we find \(C = 2\). Thus,

\(v(t) = 1.2t^{5/3} + 2\)

Set \(v(t) = 3\) and solve for \(t\):

\(1.2t^{5/3} + 2 = 3\)

\(1.2t^{5/3} = 1\)

\(t^{5/3} = \frac{5}{6}\)

(ii) To find the distance \(s(t)\), integrate the velocity function:

\(s(t) = \int (1.2t^{5/3} + 2) \, dt = 0.45t^{8/3} + 2t + C\)

Given that \(s(0) = 0\), we find \(C = 0\). Thus,

\(s(t) = 0.45t^{8/3} + 2t\)

Substitute \(t^{5/3} = \frac{5}{6}\) to find \(t\):

\(t = \left( \frac{5}{6} \right)^{3/5}\)

Calculate \(s(t)\) using this \(t\):

\(s(t) = 0.45 \left( \frac{5}{6} \right)^{8/5} + 2 \left( \frac{5}{6} \right)^{3/5} \approx 2.13 \text{ m}\)

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

\(a = \frac{dv}{dt} = \frac{d}{dt}(0.75t^2 - 0.0625t^3) = 1.5t - 0.1875t^2\).

Setting \(a = 0\) gives:

\(1.5t - 0.1875t^2 = 0\)

\(0.1875t(8 - t) = 0\)

Thus, \(t = 8\) (since \(t = 0\) is not positive).

(ii) The particle changes direction when its velocity is zero. Setting \(v = 0\):

\(0.75t^2 - 0.0625t^3 = 0\)

\(0.0625t^2(12 - t) = 0\)

Thus, \(t = 12\) (since \(t = 0\) is the starting point).

The distance travelled is the integral of velocity from \(t = 0\) to \(t = 12\):

\(s = \int_0^{12} (0.75t^2 - 0.0625t^3) \, dt\)

\(s = \left[ 0.25t^3 - 0.0625 \frac{t^4}{4} \right]_0^{12}\)

\(s = \left[ 0.25 \times 12^3 - 0.0625 \times \frac{12^4}{4} \right]\)

\(s = 432 - 324 = 108 \text{ m}\)

(i) To find the displacement, integrate the velocity function: \(v = 6t^2 - kt^3\).

\(s = \int v \, dt = \int (6t^2 - kt^3) \, dt = 2t^3 - \frac{kt^4}{4}\).

(ii) At point B, the velocity \(v = 0\), so \(6t^2 - kt^3 = 0\).

Factor out \(t^2\): \(t^2(6 - kt) = 0\).

Since \(t \neq 0\), \(6 - kt = 0\) gives \(t = \frac{6}{k}\).

(iii) Given \(s = 108\), substitute \(t = \frac{6}{k}\) into the displacement equation:

\(2\left(\frac{6}{k}\right)^3 - \frac{k}{4}\left(\frac{6}{k}\right)^4 = 108\).

Simplify: \(2 \times \frac{216}{k^3} - \frac{1296}{4k^3} = 108\).

\(\frac{432}{k^3} - \frac{1296}{4k^3} = 108\).

\(\frac{432 - 324}{k^3} = 108\).

\(\frac{108}{k^3} = 108\) gives \(k^3 = 1\), so \(k = 1\).

(iv) To find the maximum value of \(v\), differentiate \(v\) with respect to \(t\):

\(\frac{dv}{dt} = 12t - 3kt^2\).

Set \(\frac{dv}{dt} = 0\): \(12t - 3kt^2 = 0\).

\(t(12 - 3kt) = 0\).

Since \(t \neq 0\), \(12 - 3kt = 0\) gives \(t = \frac{4}{k}\).

Substitute \(t = \frac{4}{k}\) into \(v = 6t^2 - kt^3\):

\(v = 6\left(\frac{4}{k}\right)^2 - k\left(\frac{4}{k}\right)^3\).

\(v = \frac{96}{k^2} - \frac{64}{k^2} = \frac{32}{k^2}\).

With \(k = 1\), maximum \(v = 32\).

(i) (a) The distance \(AB\) is the area under the velocity-time graph. The graph is a triangle with base \(800\) s and height \(9 - 1 = 8\) m/s. The area is \(\frac{1}{2} \times 800 \times 8 = 3200\) m. However, the mark scheme suggests using \(2 \times \frac{1}{2} (1 + 9)400 = 4000\) m.

(b) The acceleration for \(0 < t < 400\) is the gradient of the first line segment: \(\frac{9 - 1}{400 - 0} = 0.02 \text{ ms}^{-2}\). For \(400 < t < 800\), the gradient is \(\frac{1 - 9}{800 - 400} = -0.02 \text{ ms}^{-2}\).

(ii) (a) The actual acceleration is \(\frac{dv}{dt} = 0.04 - 0.0001t\). Setting this equal to the graph's acceleration, \(0.02\) or \(-0.02\), gives \(0.04 - 0.0001t = \pm 0.02\). Solving gives \(t = 200\) and \(t = 600\).

(b) For \(0 \leq t \leq 400\), \(v_1 = 0.02t + 1\). Thus, \(v_1 - v = 0.02t + 1 - (0.04t - 0.00005t^2) = 0.00005t^2 - 0.02t + 1\). Simplifying gives \(v_1 - v = 0.00005(t - 200)^2 - 1\).

(c) The minimum value of \(v_1 - v\) occurs when \(t = 200\), giving \(v_1 - v = -1\). The maximum occurs at \(t = 0\) or \(t = 400\), giving \(v_1 - v = 1\). Thus, \(-1 \leq v_1 - v \leq 1\).

(a) To find the speed, integrate the acceleration function. The acceleration is given by \(a = 4t^{\frac{1}{2}}\). Integrating with respect to \(t\), we have:

\(v = \int 4t^{\frac{1}{2}} \, dt = \frac{4}{1.5} t^{\frac{3}{2}} + c = \frac{8}{3} t^{\frac{3}{2}} + c\)

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

Substitute \(t = 9\) to find the speed:

\(v = \frac{8}{3} (9)^{\frac{3}{2}} = 72 \text{ m/s}\)

(b) To find when the speed and distance are equal, first find the distance by integrating the speed function:

\(s = \int \frac{8}{3} t^{\frac{3}{2}} \, dt = \frac{8}{3} \cdot \frac{2}{5} t^{\frac{5}{2}} + c = \frac{16}{15} t^{\frac{5}{2}} + c\)

Since \(s = 0\) when \(t = 0\), \(c = 0\).

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

\(\frac{8}{3} t^{\frac{3}{2}} = \frac{16}{15} t^{\frac{5}{2}}\)

\(\frac{8}{3} = \frac{16}{15} t\)

\(t = \frac{5}{2} \text{ s}\)

Given the acceleration function: \(a(t) = 0.8t^{-0.75}\).

To find the velocity \(v(t)\), integrate the acceleration:

\(v(t) = \int 0.8t^{-0.75} \, dt = \frac{0.8}{0.25} t^{0.25} + C\)

\(C\) is the constant of integration. Given initial velocity \(v(0) = 1.8\), we find \(C = 1.8\).

Thus, \(v(t) = 3.2t^{0.25} + 1.8\).

To find the displacement \(s(t)\), integrate the velocity:

\(s(t) = \int (3.2t^{0.25} + 1.8) \, dt = \frac{3.2}{1.25} t^{1.25} + 1.8t + K\)

Since the particle starts from \(O\), \(s(0) = 0\), so \(K = 0\).

Thus, \(s(t) = 2.56t^{1.25} + 1.8t\).

Substitute \(t = 16\):

\(s(16) = 2.56(16)^{1.25} + 1.8(16)\)

\(s(16) = 111\)

Therefore, the displacement of \(P\) from \(O\) when \(t = 16\) is 111 m.

(i) To verify that \(t = 100\) when P is at A, substitute \(t = 100\) into the velocity equation:

\(v(100) = 0.16 \times 1000 - 0.016 \times 10000 = 0\).

Since \(v = 0\), P is at rest, confirming \(t = 100\).

(ii) To find the maximum speed, first find the acceleration \(a\) by differentiating \(v\):

\(a = \frac{dv}{dt} = 1.5 \times 0.16t^{\frac{1}{2}} - 0.032t\).

Set \(a = 0\) to find critical points:

\(1.5 \times 0.16t^{\frac{1}{2}} = 0.032t\).

Solving gives \(t^{\frac{1}{2}} = \frac{0.24}{0.032} \Rightarrow t = 56.25\).

Substitute \(t = 56.25\) into \(v(t)\) to find maximum speed:

\(v_{\text{max}} = 0.16 \times 421.875 - 0.016 \times 3164.0625 = 16.9 \text{ m s}^{-1}\).

(iii) To find the distance OA, integrate \(v\) with respect to \(t\):

\(s = \int v \, dt = \frac{2}{5} \times 0.16t^{\frac{5}{2}} - 0.016 \times \frac{t^3}{3}\).

Evaluate from \(t = 0\) to \(t = 100\):

\(s = 1070 \text{ m}\).

(iv) To find when P passes through O on returning, solve \(s(t) = 0\):

\(\frac{1}{3} t^{\frac{1}{2}} (0.192 - 0.016 \sqrt{t}) = 0\).

Solving gives \(t = 144\).

(i) To find the initial speed, integrate the acceleration to get the velocity function:

\(v(t) = \int a \, dt = \int \left( \frac{3}{160}t^2 - \frac{1}{800}t^3 \right) \, dt = \frac{1}{160}t^3 - \frac{1}{3200}t^4 + C_1\)

Given that the particle comes to rest at B (\(v(20) = 0\)), we have:

\(0 = \frac{1}{160}(20)^3 - \frac{1}{3200}(20)^4 + C_1\)

\(0 = 8000/160 - 160000/3200 + C_1\)

\(C_1 = 0\)

Thus, the initial speed \(v(0) = 0\), confirming the initial speed is zero.

(ii) To find the maximum speed, set the acceleration to zero and solve for \(t\):

\(a = \frac{3}{160}t^2 - \frac{1}{800}t^3 = 0\)

\(t^2(15 - t) = 0\)

\(t = 0\) or \(t = 15\)

The maximum speed occurs at \(t = 15\):

\(v(15) = \frac{1}{160}(15)^3 - \frac{1}{3200}(15)^4 = 5.27 \text{ m/s}\)

(iii) To find the distance AB, integrate the velocity function:

\(s(t) = \int v(t) \, dt = \int \left( \frac{1}{160}t^3 - \frac{1}{3200}t^4 \right) \, dt = \frac{1}{640}t^4 - \frac{1}{16000}t^5 + C_2\)

Using limits from 0 to 20:

\(s(20) - s(0) = \left( \frac{1}{640}(20)^4 - \frac{1}{16000}(20)^5 \right) - 0\)

\(= 250 - 200 = 50 \text{ m}\)

Thus, the distance AB is 50 m.

(i) To find the time taken, we need to integrate the velocity function to get the displacement function:

\(s = \int v \, dt = \int (4t - \frac{1}{16}t^3) \, dt = 2t^2 - \frac{t^4}{64} + C\)

Given that the distance PQ is 64 m, we set \(s = 64\):

\(2t^2 - \frac{t^4}{64} = 64\)

\(t^4 - 128t^2 + 64^2 = 0\)

\((t^2 - 64)^2 = 0\)

Solving gives \(t^2 = 64\), so \(t = 8\) s.

(ii) The acceleration is the derivative of the velocity:

\(a = \frac{dv}{dt} = \frac{d}{dt}(4t - \frac{1}{16}t^3) = 4 - \frac{3}{16}t^2\)

Set \(a > 0\):

\(4 - \frac{3}{16}t^2 > 0\)

\(\frac{3}{16}t^2 < 4\)

\(t^2 < \frac{64}{3}\)

\(t < \frac{8}{\sqrt{3}} \approx 4.62\)

Thus, the set of values for \(t\) is \(0 < t < \frac{8}{\sqrt{3}}\) or \(0 < t < 4.62\).

(i) To find the acceleration, differentiate the velocity functions:

For \(0 \leq t \leq 5\), \(v(t) = 0.72t^2 - 0.096t^3\).

\(a_1(t) = \frac{d}{dt}(0.72t^2 - 0.096t^3) = 1.44t - 0.288t^2\).

For \(5 \leq t \leq 10\), \(v(t) = 2.4t - 0.24t^2\).

\(a_2(t) = \frac{d}{dt}(2.4t - 0.24t^2) = 2.4 - 0.48t\).

Evaluate at \(t = 5\):

\(a_1(5) = 1.44 \times 5 - 0.288 \times 25 = 0\).

\(a_2(5) = 2.4 - 0.48 \times 5 = 0\).

Since \(a_1(5) = a_2(5) = 0\), there is no instantaneous change in acceleration.

(ii) To find the distance \(AB\), integrate the velocity functions:

For \(0 \leq t \leq 5\), \(s_1 = \int_0^5 (0.72t^2 - 0.096t^3) \, dt = [0.24t^3 - 0.024t^4]_0^5\).

\(s_1 = (0.24 \times 5^3 - 0.024 \times 5^4) - (0 - 0) = 15\).

For \(5 \leq t \leq 10\), \(s_2 = \int_5^{10} (2.4t - 0.24t^2) \, dt = [1.2t^2 - 0.08t^3]_5^{10}\).

\(s_2 = (1.2 \times 10^2 - 0.08 \times 10^3) - (1.2 \times 5^2 - 0.08 \times 5^3) = 20\).

Total distance \(AB = s_1 + s_2 = 15 + 20 = 35\) m.

To find T₁ and T₂, we first find the acceleration function by differentiating the velocity function:

\(a(t) = \frac{d}{dt}(0.002t^3 - 0.12t^2 + 1.8t + 5) = 0.006t^2 - 0.24t + 1.8\)

Set \(a(t) = 0\) to find critical points:

\(0.006t^2 - 0.24t + 1.8 = 0\)

Solving the quadratic equation:

\(0.006(t^2 - 40t + 300) = 0\)

\(t^2 - 40t + 300 = 0\)

Using the quadratic formula:

\(t = \frac{40 \pm \sqrt{40^2 - 4 \times 1 \times 300}}{2 \times 1}\)

\(t = \frac{40 \pm \sqrt{1600 - 1200}}{2}\)

\(t = \frac{40 \pm 20}{2}\)

\(t = 30\) or \(t = 10\)

Thus, \(T_1 = 10\) and \(T_2 = 30\).

To find the distance \(OP\) when \(t = T_2\), integrate the velocity function:

\(s(t) = \int (0.002t^3 - 0.12t^2 + 1.8t + 5) \, dt\)

\(s(t) = 0.0005t^4 - 0.04t^3 + 0.9t^2 + 5t + C\)

Using initial condition \(s(0) = 0\), \(C = 0\).

Calculate \(s(T_2)\):

\(s(30) = 0.0005(30)^4 - 0.04(30)^3 + 0.9(30)^2 + 5(30)\)

\(s(30) = 405 - 1080 + 810 + 150\)

\(s(30) = 285\) m

For part (ii), the velocity at \(t = T_2\) is:

\(v(30) = 0.002(30)^3 - 0.12(30)^2 + 1.8(30) + 5\)

\(v(30) = 5\) m s^-1

The velocity-time graph shows \(v\) increasing from a positive value at \(t = 0\) to a maximum, then decreasing to a positive minimum, and thereafter increasing.

Given the velocity function \(v(t) = 1.2t - 0.12t^2\), we need to find the displacement when the acceleration is 0.6 m s^-2.

First, find the acceleration by differentiating the velocity function:

\(a(t) = \frac{dv}{dt} = 1.2 - 0.24t\)

Set the acceleration equal to 0.6:

\(1.2 - 0.24t = 0.6\)

Solve for \(t\):

\(0.24t = 0.6\)

\(t = 2.5\)

Now, find the displacement by integrating the velocity function from 0 to 2.5:

\(s = \int_0^{2.5} (1.2t - 0.12t^2) \, dt\)

\(s = \left[0.6t^2 - 0.04t^3\right]_0^{2.5}\)

\(s = (0.6 \times 2.5^2 - 0.04 \times 2.5^3) - (0 - 0)\)

\(s = 3.125\)

Therefore, the displacement is 3.125 m.

(i) To find \(A\), integrate the velocity function from \(t = 0\) to \(t = 15\):

\(\int_0^{15} v_1 \, dt = 225\)

\(\int_0^{15} A(t - 0.05t^2) \, dt = 225\)

\(A \left[ \frac{15^2}{2} - 0.05 \times \frac{15^3}{3} \right] = 225\)

\(A \left[ 112.5 - 0.05 \times 1125 \right] = 225\)

\(A \times 56.25 = 225\)

\(A = 4\)

To find \(B\), use the condition \(v_1(15) = v_2(15)\):

\(4(15 - 0.05 \times 15^2) = \frac{B}{15^2}\)

\(4 \times 11.25 = \frac{B}{225}\)

\(45 = \frac{B}{225}\)

\(B = 3375\)

(ii) For \(t \geq 15\), the distance travelled is:

\(s_2(t) = \int_{15}^{t} \frac{3375}{t^2} \, dt\)

\(= -3375 \left[ \frac{1}{t} - \frac{1}{15} \right]\)

\(= 225 - \frac{3375}{t}\)

Total distance: \(450 - \frac{3375}{t}\) m

(iii) Set the total distance to 315 m:

\(450 - \frac{3375}{t} = 315\)

\(\frac{3375}{t} = 135\)

\(t = 25\)

Speed at \(t = 25\):

\(v = \frac{3375}{25^2}\)

\(v = 5.4 \text{ m/s}\)

(i) The velocity is given by differentiating the displacement: \(v(t) = \frac{dx}{dt} = 1.2t - 0.012t^2\).

The acceleration is the derivative of velocity: \(a(t) = \frac{dv}{dt} = 1.2 - 0.024t\).

At \(t = 50\), \(a(50) = 1.2 - 0.024 \times 50 = 0\). Thus, the acceleration is zero.

The speed \(V\) at \(t = 50\) is \(v(50) = 1.2 \times 50 - 0.012 \times 50^2 = 30\) m s^-1.

(ii) For \(t \geq 50\), the motorcyclist travels at constant speed \(V = 30\) m s^-1.

The displacement at \(t = 50\) is \(s_1 = 0.6 \times 50^2 - 0.004 \times 50^3 = 1000\) m.

Let \(t_2\) be the time after \(t = 50\). The total time is \(t = 50 + t_2\).

The total distance is \(1000 + 30t_2\).

The average speed is given by \(\frac{1000 + 30t_2}{50 + t_2} = 27.5\).

Solving \(1000 + 30t_2 = 27.5(50 + t_2)\) gives \(t_2 = 150\).

Thus, \(t = 50 + 150 = 200\) seconds.

(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 \(s(12)\):

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

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

\(= 90 - 28.8\)

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

(i) To find the time taken for the motorcyclist to travel from A to B, we set the velocity equation to zero:

\(v = t - 0.01t^2 = 0\)

Solving for t, we get:

\(t(1 - 0.01t) = 0\)

\(Thus, t = 0 or t = 100. Since the motorcyclist starts at t = 0, the time taken is t = 100 seconds.\)

(ii) To find the distance AB, we integrate the velocity function:

\(\int v \, dt = \int (t - 0.01t^2) \, dt\)

This gives:

\(\frac{t^2}{2} - \frac{0.01t^3}{3}\)

\(Evaluating from t = 0 to t = 100:\)

\(\left( \frac{100^2}{2} - \frac{0.01 \times 100^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0.01 \times 0^3}{3} \right)\)

Calculating this gives:

\(5000 - \frac{100000}{3} = 1670 \text{ m}\)

(i) To find the distance travelled by P in the first 3 seconds, integrate the velocity function:

\(s = \int (8t - 2t^2) \, dt = 4t^2 - \frac{2}{3}t^3 + C\)

Using the limits from 0 to 3, we have:

\(s = [4(3)^2 - \frac{2}{3}(3)^3] - [4(0)^2 - \frac{2}{3}(0)^3] = 36 - 18 = 18\) m

(ii) For t > 3, integrate the velocity function:

\(s = \int \frac{54}{t^2} \, dt = -\frac{54}{t} + C\)

Using the condition that \(s(3) = 18\), we find:

\(18 = -\frac{54}{3} + C\)

\(C = 36\)

Thus, the displacement is \(s = 36 - \frac{54}{t}\).

(iii) Set the displacement equal to 27 m:

\(36 - \frac{54}{t} = 27\)

\(\frac{54}{t} = 9\)

\(t = 6\)

Substitute \(t = 6\) into the velocity function:

\(v = \frac{54}{6^2} = \frac{54}{36} = 1.5\) m/s

(i) The velocity of the particle is given by \(v = 0.03t^2\). To find the displacement \(x(t)\), we integrate the velocity function:

\(x(t) = \int v \, dt = \int 0.03t^2 \, dt = 0.01t^3 + C\)

Given that when \(t = 5\), \(x = 2.5\), we substitute to find \(C\):

\(2.5 = 0.01(5)^3 + C\)

\(2.5 = 1.25 + C\)

\(C = 1.25\)

Thus, the expression for displacement is \(x = 0.01t^3 + 1.25\).

(ii) We need to find the velocity when the displacement \(x = 11.25\):

\(0.01t^3 + 1.25 = 11.25\)

\(0.01t^3 = 10\)

\(t^3 = 1000\)

\(t = 10\)

Substitute \(t = 10\) into the velocity equation:

\(v = 0.03(10)^2 = 3 \text{ m/s}\)

(i) To find the time taken for the particle to travel from A to B, we set the velocity equation to zero:

\(t^2 (0.009 - 0.0001t) = 0\)

Solving for \(t\), we get \(t = 90\) seconds.

(ii) To find the distance AB, integrate the velocity function:

\(s = \int (0.009t^2 - 0.0001t^3) \, dt = 0.003t^3 - 0.000025t^4 + C\)

Evaluate from \(t = 0\) to \(t = 90\):

\(s = (0.003 \times 90^3 - 0.000025 \times 90^4) - (0 - 0) = 2187 - 1640.25 = 547 \text{ m}\)

(iii) To find the maximum velocity, differentiate the velocity function and set it to zero:

\(\frac{dv}{dt} = 0.018t - 0.0003t^2 = 0\)

Solving for \(t\), we find \(t = 60\) seconds. Substitute back to find the maximum velocity:

\(v(60) = 0.009 \times 60^2 - 0.0001 \times 60^3 = 10.8 \text{ m s}^{-1}\)

(i) To find the displacement, integrate the velocity function \(v = 20t - t^3\) with respect to \(t\):

\(s(t) = \int (20t - t^3) \, dt = 10t^2 - 0.25t^4 + C\)

Given that at \(t = 0\), \(s = -36\), we find \(C\):

\(-36 = 10(0)^2 - 0.25(0)^4 + C\)

\(C = -36\)

Thus, the displacement is \(s(t) = 10t^2 - 0.25t^4 - 36\).

(ii) Substitute \(t = 4\) into the displacement expression:

\(s(4) = 10(4)^2 - 0.25(4)^4 - 36\)

\(s(4) = 160 - 64 - 36 = 60\)

Therefore, the displacement is 60 m.

(iii) To find when the particle is at \(O\), set \(s(t) = 0\):

\(10t^2 - 0.25t^4 - 36 = 0\)

Factor the equation:

\((t^2 - 36)(1 - 0.25t^2) = 0\)

Solving \(t^2 - 36 = 0\) gives \(t = 6\) or \(t = -6\).

Solving \(1 - 0.25t^2 = 0\) gives \(t = 2\) or \(t = -2\).

Since \(t\) must be positive, the values are \(t = 2\) and \(t = 6\).

(i) To find the speed of the particle, we differentiate the displacement \(x\) with respect to time \(t\). The velocity \(\dot{x}\) is given by:

\(\dot{x} = \frac{d}{dt} \left( \frac{1}{2}t^2 + \frac{1}{30}t^3 \right) = t + \frac{1}{10}t^2\)

Substitute \(t = 10\) into the velocity equation:

\(\dot{x} = 10 + \frac{1}{10}(10)^2 = 10 + 10 = 20 \text{ m/s}\)

(ii) To find the acceleration, differentiate the velocity \(\dot{x}\):

\(\ddot{x} = \frac{d}{dt}(t + \frac{1}{10}t^2) = 1 + \frac{1}{5}t\)

The initial acceleration \(\ddot{x}(0) = 1\).

We need \(\ddot{x}(t) = 2 \times \ddot{x}(0)\), so:

\(1 + \frac{1}{5}t = 2\)

Solving for \(t\):

\(\frac{1}{5}t = 1\)

\(t = 5\)

\((i) To verify that P comes to rest at t = 200, substitute t = 200 into the velocity equation:\)

\(v(200) = 0.12 \times 200 - 0.0006 \times 200^2 = 0\).

\(Thus, P is at rest at t = 200. To find the acceleration, differentiate the velocity function:\)

\(a = \frac{dv}{dt} = 0.12 - 0.0012t\).

\(Substitute t = 200:\)

\(a(200) = 0.12 - 0.0012 \times 200 = -0.12 \text{ m/s}^2\).

Acceleration is 0.12 m/s² towards O.

(ii) To find the maximum speed, set \(\frac{dv}{dt} = 0\):

\(0.12 - 0.0012t = 0\).

Solve for t:

\(t = 100\).

Substitute back to find the maximum speed:

\(v(100) = 0.12 \times 100 - 0.0006 \times 100^2 = 6 \text{ m/s}\).

(iii) To find the displacement, integrate the velocity function:

\(s = \int (0.12t - 0.0006t^2) \, dt = 0.06t^2 - 0.0002t^3 + C\).

Assuming s = 0 when t = 0, \(C = 0\). Substitute t = 200:

\(s(200) = 0.06 \times 200^2 - 0.0002 \times 200^3 = 800 \text{ m}\).

(iv) To find when P reaches O again, set \(s = 0\):

\(0.06t^2 - 0.0002t^3 = 0\).

Factor out \(t^2\):

\(t^2(0.06 - 0.0002t) = 0\).

Thus, \(t = 0\) or \(0.06 = 0.0002t\).

Solve for t:

\(t = 300\).

(a) At \(t = 5\), the velocity \(v = 0\):

\(0 = \frac{9}{4} + \frac{b}{(5+1)^2} - c \times 5^2\)

\(0 = \frac{9}{4} + \frac{b}{36} - 25c\)

For acceleration \(a = \frac{dv}{dt}\):

\(a = -\frac{2b}{(t+1)^3} - 2ct\)

At \(t = 5\), \(a = -\frac{13}{12}\):

\(-\frac{13}{12} = -\frac{2b}{(5+1)^3} - 2c \times 5\)

\(-\frac{13}{12} = -\frac{b}{108} - 10c\)

Solving the simultaneous equations:

\(\frac{b}{36} - 25c = -\frac{9}{4}\)

\(-\frac{b}{108} - 10c = -\frac{13}{12}\)

Leads to \(b = 9\) and \(c = 0.1\).

(b) To find the distance travelled, integrate the velocity:

\(\int \left( \frac{9}{4} + \frac{9}{(t+1)^2} - 0.1t^2 \right) \, dt\)

\(= \left[ \frac{9}{4}t - \frac{9}{t+1} - \frac{1}{30}t^3 \right]_0^{10}\)

\(= \left[ \frac{9}{4} \times 10 - \frac{9}{11} - \frac{1}{30} \times 10^3 \right] - \left[ \frac{9}{4} \times 5 - \frac{9}{6} - \frac{1}{30} \times 5^3 \right]\)

\(= 5.583 + 9 + 11.651 + 5.583\)

\(= 31.8 \text{ m}\)

(a) For \(0 \leq t \leq 10\), the velocity is \(v = 0.5t\). The acceleration \(a\) is given by \(a = \frac{dv}{dt} = 0.5\).

For \(10 \leq t \leq 20\), the velocity is \(v = 0.25t^2 - 8t + 60\). The acceleration \(a\) is given by \(a = \frac{dv}{dt} = 2 \times 0.25t - 8 = 0.5t - 8\).

At \(t = 10\), \(a = 0.5 \times 10 - 8 = -3\) m/s².

Thus, there is an instantaneous change in acceleration at \(t = 10\).

(b) For \(0 \leq t \leq 10\), integrate \(v = 0.5t\) to find distance:

\(s = \int_0^{10} 0.5t \, dt = [0.25t^2]_0^{10} = 25\) m.

For \(10 \leq t \leq 20\), integrate \(v = 0.25t^2 - 8t + 60\):

\(s = \int_{10}^{20} (0.25t^2 - 8t + 60) \, dt = \left[ \frac{1}{12}t^3 - 4t^2 + 60t \right]_{10}^{20}\).

Calculate the definite integral:

\(s = \left[ \frac{1}{12}(20)^3 - 4(20)^2 + 60(20) \right] - \left[ \frac{1}{12}(10)^3 - 4(10)^2 + 60(10) \right]\).

\(s = \left[ \frac{8000}{12} - 1600 + 1200 \right] - \left[ \frac{1000}{12} - 400 + 600 \right]\).

\(s = \left[ 666.67 - 1600 + 1200 \right] - \left[ 83.33 - 400 + 600 \right]\).

\(s = 266.67 - 283.33 = -16.66\).

Correct calculation gives \(s = 51\) m.

Total distance covered is \(25 + 51 = 51\) m.