(a) To find when the acceleration is zero, differentiate the velocity function:

\(a = \frac{dv}{dt} = 3(-0.1)t^2 + 2(1.8)t - 6 = -0.3t^2 + 3.6t - 6\).

Set \(a = 0\):

\(-0.3t^2 + 3.6t - 6 = 0\).

Solving the quadratic equation gives \(t = 2\) and \(t = 10\). Thus, \(p = 2\) and \(q = 10\).

(b) The velocity at \(t = 2\) is:

\(v = -0.1(2)^3 + 1.8(2)^2 - 6(2) + 5.6 = 0\) m/s.

The velocity at \(t = 10\) is:

\(v = -0.1(10)^3 + 1.8(10)^2 - 6(10) + 5.6 = 25.6\) m/s.

The velocity-time graph is a curve with a single minimum turning point followed by a single maximum turning point.

(c) To find the total distance, integrate the velocity function:

\(s = \int v \, dt = \int (-0.1t^3 + 1.8t^2 - 6t + 5.6) \, dt\).

\(s = -0.025t^4 + 0.6t^3 - 3t^2 + 5.6t + C\).

Calculate the distance from \(t = 0\) to \(t = 14\):

\(s = 176.4\) m.

Calculate the distance from \(t = 14\) to \(t = 15\):

\(s = 8.025\) m.

\(Total distance = 176.4 + 8.025 = 184.425\) m.\)

(i) For \(15 \leq t \leq 35\), the velocity is \(v = a + bt^2\). At \(t = 35\), \(v = 0\), so:

\(0 = a + b \times 35^2\)

\(0 = a + 1225b\)

At \(t = 15\), \(v = 2 \times 15 + 10 = 40\), so:

\(40 = a + b \times 15^2\)

\(40 = a + 225b\)

Solving these equations:

\(1000b = -40 \rightarrow b = -0.04\)

\(a = 0.04 \times 352 = 49\)

(ii) The velocity-time graph consists of:

• \(0 \leq t \leq 5\): Increasing quadratic from (0,0) to (5,20), concave up.
• \(5 \leq t \leq 15\): Line from (5,20) to (15,40).
• \(15 \leq t \leq 35\): Decreasing quadratic from (15,40) to (35,0), concave down.

(iii) Total distance:

\(A_1 = \int_0^5 0.8t^2 \, dt = \frac{100}{3}\)

\(A_2 = \frac{1}{2} (20 + 40) \times 10 = 300\)

\(A_3 = \int_{15}^{35} (a + bt^2) \, dt = 49t - \frac{0.04}{3}t^3 \bigg|_{15}^{35}\)

\(A_3 = 453.3333 = \frac{1360}{3}\)

Total Distance = \(\frac{100}{3} + 300 + \frac{1360}{3} = 787 \text{ m}\)

(i) To find \(k\), consider the velocity at \(t = 4\) from the first equation: \(v = 5t(t - 2)\). Substituting \(t = 4\), we get:

\(v = 5 \times 4 \times (4 - 2) = 40\).

Thus, \(k = 40\).

(ii) The velocity-time graph consists of three segments:

• For \(0 \leq t \leq 4\), the graph is a quadratic curve \(v = 5t(t - 2)\) with a minimum at \(t = 1\) and \(v = 0\) at \(t = 2\).
• For \(4 \leq t \leq 14\), the graph is a horizontal line at \(v = 40\).
• For \(14 \leq t \leq 20\), the graph is a line with a negative gradient \(v = 68 - 2t\).

(iii) For \(0 \leq t \leq 4\), the acceleration \(a = \frac{dv}{dt} = \frac{d}{dt}(5t(t - 2)) = 10t - 10\).

The acceleration is positive when \(10t - 10 > 0\), which gives \(t > 1\).

Thus, the set of values of \(t\) for which the acceleration is positive is \(1 < t < 4\).

(iv) To find the total distance travelled:

For \(0 \leq t \leq 4\), integrate \(v = 5t(t - 2)\):

\(\int (5t^2 - 10t) \, dt = \left[ \frac{5}{3}t^3 - 5t^2 \right]_0^4 = \frac{100}{3}\).

For \(4 \leq t \leq 14\), the distance is \(40 \times 10 = 400\).

For \(14 \leq t \leq 20\), integrate \(v = 68 - 2t\):

\(\int (68 - 2t) \, dt = \left[ 68t - t^2 \right]_{14}^{20} = 204\).

Total distance = \(\frac{100}{3} + 400 + 204 = 644\) m.

(i) To show that P is at rest, we need \(v(t) = 0\). The velocity function is \(v(t) = 0.0001t^3 - 0.015t^2 + 0.5t\). Factorizing gives \(0.0001t(t - 50)(t - 100) = 0\). Thus, \(v(t) = 0\) when \(t = 0, 50, 100\).

(ii) The acceleration \(a(t)\) is the derivative of \(v(t)\): \(a(t) = \frac{dv}{dt} = 0.0003t^2 - 0.03t + 0.5\). Setting \(a(t) = 0\) gives \(0.0003t^2 - 0.03t + 0.5 = 0\). Solving this quadratic equation, \(t^2 - 100t + 1667 = 0\), we find \(t = 21.1\) and \(t = 78.9\). The velocities are \(v(21.1) = 4.81\) and \(v(78.9) = -4.81\). The velocity-time graph is a convex curve from \((0,0)\) to \((50,0)\) with \(v > 0\) and a maximum point. The curve from \((50,0)\) to \((100,0)\) is the first curve rotated 180° about \((50,0)\).

(iii) To find the greatest distance, integrate \(v(t)\) to get the displacement \(s(t)\): \(s(t) = 0.000025t^4 - 0.005t^3 + 0.25t^2 + c\). Using limits \(t = 0\) and \(t = 50\), the greatest distance is \(156\) m.

(i) To find the velocity immediately before the particle hits the wall at \(t = 60\), substitute \(t = 60\) into \(v = 0.05t - 0.0005t^2\):

\(v = 0.05(60) - 0.0005(60)^2 = 3 - 1.8 = 1.2 \text{ ms}^{-1}\).

Immediately after hitting the wall, the velocity is given as \(-1 \text{ ms}^{-1}\).

(ii) The total distance travelled is the sum of the distances from \(O\) to the wall \(W\) and from \(W\) to \(A\).

Distance \(OW = \int_{0}^{60} (0.05t - 0.0005t^2) \, dt = \left[ 0.025t^2 - 0.0005 \frac{t^3}{3} \right]_{0}^{60} = 54 \text{ m}\).

Distance \(WA = \int_{60}^{100} (0.025t - 2.5) \, dt = \left[ 0.0125t^2 - 2.5t \right]_{60}^{100} = 20 \text{ m}\).

Total distance = \(54 + 20 = 74 \text{ m}\).

(iii) To find the maximum speed, consider the velocity function \(v = 0.05t - 0.0005t^2\) for \(0 \leq t < 60\). Differentiate to find the critical points:

\(\frac{dv}{dt} = 0.05 - 0.001t = 0 \Rightarrow t = 50\).

Substitute \(t = 50\) into the velocity equation:

\(v = 0.05(50) - 0.0005(50)^2 = 1.25 \text{ ms}^{-1}\).

The maximum speed is \(1.25 \text{ ms}^{-1}\) at \(t = 50\).

The velocity-time graph is a quadratic curve from \((0,0)\) to \((60,1.2)\) with a maximum at \((50,1.25)\), followed by a straight line from \((60,-1)\) to \((100,0)\).