(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 \((100,0)\).

(i) To find the distance AB, set the derivative of the distance function to zero to find when the car stops:

\(\frac{d}{dt}[0.0000117(400t^3 - 3t^4)] = 0\)

\(0.0000117(1200t^2 - 12t^3) = 0\)

\(1200t^2 = 12t^3\)

\(t = 0, 100\)

Distance \(AB = 0.0000117(400(100)^3 - 3(100)^4) = 1170\) m.

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

\(v(t) = \frac{d}{dt}[0.0000117(400t^3 - 3t^4)] = 0.0000117(1200t^2 - 12t^3)\)

\(\frac{d}{dt}[v(t)] = 0.0000117(2400t - 36t^2) = 0\)

\(2400t = 36t^2\)

\(t = 0, \frac{200}{3}\)

Maximum speed \(v_{\text{max}} = 0.0000117(1200(\frac{200}{3})^2 - 12(\frac{200}{3})^3) = 20.8\) m/s.

(iii) (a) At A, \(a(t) = 0\).

(b) At B, \(a(t) = 0.0000117(2400 \times 100 - 36 \times 100^2) = -1.40\) m/s².

(iv) The velocity-time graph starts at zero, increases to a maximum, and then decreases back to zero. The maximum is closer to \(t = 100\) than \(t = 0\). The graph has a zero gradient at \(t = 0\) and an inflection point closer to \(t = 0\) than \(t = 100\).

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

\(v = \int (12 - 2t) \, dt = 12t - \frac{2t^2}{2} + c = 12t - t^2 + c\)

Using the initial condition \(v(0) = -20\), we find \(c = -20\).

Thus, \(v = 12t - t^2 - 20\).

Set \(v = 0\) to find when P is at rest:

\(12t - t^2 - 20 = 0\)

\(t^2 - 12t + 20 = 0\)

Solving this quadratic equation gives \(t = 2\) and \(t = 10\).

The velocity-time graph is an inverted quadratic starting at \((0, -20)\) and ending at \((12, -20)\), with roots at \(t = 2\) and \(t = 10\).

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

\(s = \int_{0}^{12} (12t - t^2 - 20) \, dt\)

\(s = \left[ \frac{12t^2}{2} - \frac{t^3}{3} - 20t \right]_{0}^{12}\)

Calculate the definite integral in segments:

\(s = \left[ \frac{12(2)^2}{2} - \frac{(2)^3}{3} - 20(2) \right] + \left[ \frac{12(10)^2}{2} - \frac{(10)^3}{3} - 20(10) \right] - \left[ \frac{12(0)^2}{2} - \frac{(0)^3}{3} - 20(0) \right]\)

\(s = 18.7 + 85.3 + 18.7 = 123 \text{ m}\)

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

\(a = \frac{d}{dt}(pt^2 - qt) = 2pt - q.\)

Given that \(a = 0\) when \(t = 2\), we have:

\(2p(2) - q = 0 \Rightarrow 4p = q.\)

At \(t = 6\), the velocity must be continuous, so:

\(p(6)^2 - q(6) = 63 - 4.5(6).\)

Solving these equations:

\(36p - 6q = 36\)

\(4p = q\)

Substitute \(q = 4p\) into the first equation:

\(36p - 6(4p) = 36 \Rightarrow 36p - 24p = 36 \Rightarrow 12p = 36 \Rightarrow p = 3.\)

Then \(q = 4p = 12.\)

(b) The velocity-time graph consists of a quadratic curve from \(t = 0\) to \(t = 6\) and a straight line from \(t = 6\) to \(t = 14\). Key points are \((0, 0)\), \((6, 36)\), and \((14, 0)\).

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

For \(0 \leq t \leq 6\):

\(s = \int_0^6 (3t^2 - 12t) \, dt = \left[ t^3 - 6t^2 \right]_0^6 = 216 - 144 = 72.\)

For \(6 \leq t \leq 14\):

\(s = \int_6^{14} (63 - 4.5t) \, dt = \left[ 63t - 2.25t^2 \right]_6^{14} = 144.\)

\(Total distance = 72 + 144 = 216 m.\)

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

\(2t^2 - 5t + 3 = 0\).

Factorize to get \((2t - 3)(t - 1) = 0\), giving \(t = 1\) or \(t = 1.5\).

To find the minimum velocity, complete the square or use calculus. The minimum occurs at \(t = 1.25\):

\(v = 2\left(\frac{5}{4}\right)^2 - 5\left(\frac{5}{4}\right) + 3 = -0.125\) m s\(^{-1}\).

(b) The velocity-time graph is a quadratic curve with roots at \(t = 1\) and \(t = 1.5\). Key points are (1.25, -0.125), (0, 3), (1, 0), (1.5, 0), (3, 6).

(c) To find the distance, integrate the velocity function from \(t = 1\) to \(t = 1.5\):

\(s = \int_{1}^{1.5} (2t^2 - 5t + 3) \, dt\).

Calculate:

\(s = \left[ \frac{2}{3}t^3 - \frac{5}{2}t^2 + 3t \right]_{1}^{1.5}\).

\(s = \left[ \frac{2}{3}(1.5)^3 - \frac{5}{2}(1.5)^2 + 3(1.5) \right] - \left[ \frac{2}{3}(1)^3 - \frac{5}{2}(1)^2 + 3(1) \right]\).

\(s = 0.0417\) m.

(a) To find the velocity, integrate the acceleration function over each interval:

For \(0 \leq t < 2\), \(a = 6 + 4t\). Integrating gives \(v = 6t + 2t^2 + C\).

For \(2 \leq t < 4\), \(a = 14\). Integrating gives \(v = 14t + C\).

For \(4 \leq t \leq 16\), \(a = 16 - 2t\). Integrating gives \(v = 16t - t^2 + C\).

Using the condition \(v = 0\) at \(t = 16\), solve \(0 = 16(16) - 16^2 + C\) to find \(C = 0\).

Set \(v = 55\) and solve \(55 = 16t - t^2\) for \(4 \leq t \leq 16\). This gives \(t = 5\) and \(t = 11\).

(b) The velocity-time graph is a positive quadratic for \(0 \leq t < 2\) and a negative quadratic for \(4 \leq t \leq 16\), passing through the origin and \((16, 0)\).

(c) The distance travelled while decelerating is found by integrating the velocity from \(t = 8\) to \(t = 16\):

\(s = \int_{8}^{16} (16t - t^2) \, dt = \left[ 8t^2 - \frac{1}{3}t^3 \right]_{8}^{16}\)

\(s = \left( 2048 - 1365.33 \right) - \left( 512 - 170.67 \right) = 341 \frac{1}{3} \text{ m}\)

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

1. For \(0 \leq t \leq 5\), the graph is a straight line with equation \(v = 2t + 1\).
2. For \(5 \leq t \leq 7\), the graph is a downward-opening parabola with equation \(v = 36 - t^2\).
3. For \(7 \leq t \leq 13.5\), the graph is a straight line with equation \(v = 2t - 27\).

(b) To find the acceleration at \(t = 6\), differentiate the velocity function for \(5 \leq t \leq 7\):

\(v = 36 - t^2\)

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

At \(t = 6\), \(a = -2(6) = -12\).

(c) The total distance travelled is the integral of the absolute value of velocity over the given intervals:

\(s = \int_{0}^{5} (2t + 1) \, dt + \int_{5}^{6} (36 - t^2) \, dt + \int_{6}^{7} (36 - t^2) \, dt + \int_{7}^{13.5} (2t - 27) \, dt\)

Calculate each integral:

\(\int_{0}^{5} (2t + 1) \, dt = [t^2 + t]_{0}^{5} = 25 + 5 = 30\)

\(\int_{5}^{6} (36 - t^2) \, dt = [36t - \frac{t^3}{3}]_{5}^{6} = (216 - 72) - (180 - \frac{125}{3}) = 36 - 41.67 = -5.67\)

\(\int_{6}^{7} (36 - t^2) \, dt = [36t - \frac{t^3}{3}]_{6}^{7} = (252 - \frac{343}{3}) - (216 - 72) = 36 - 41.67 = -5.67\)

\(\int_{7}^{13.5} (2t - 27) \, dt = [t^2 - 27t]_{7}^{13.5} = (182.25 - 364.5) - (49 - 189) = -182.25 + 140 = -42.25\)

Total distance \(s = 30 + 5.67 + 5.67 + 42.25 = 84.25\).

(i) To find the time when the velocity is zero, integrate the acceleration to find the velocity:

\(v = \int (5.4 - 1.62t) \, dt = 5.4t - 0.81t^2 + C\).

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

Set \(v = 0\):

\(5.4t - 0.81t^2 = 0\)

\(t(5.4 - 0.81t) = 0\)

\(t = 0\) or \(t = \frac{5.4}{0.81} = \frac{20}{3}\) s.

Thus, the positive value of \(t\) is \(\frac{20}{3}\) s.

(ii) To find the velocity at \(t = 10\):

\(v(10) = 5.4(10) - 0.81(10)^2 = 54 - 81 = -27 \text{ m/s}\).

The velocity-time graph is an inverted parabola, starting at \(v = 0\) at \(t = 0\), passing through \(\left( \frac{20}{3}, 0 \right)\), and \(v = -27\) at \(t = 10\).

(iii) To find the total distance travelled, integrate the velocity to find the displacement:

\(s = \int (5.4t - 0.81t^2) \, dt = 2.7t^2 - 0.27t^3 + C\).

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

Calculate the displacement at \(t = \frac{20}{3}\):

\(s = 2.7 \left( \frac{20}{3} \right)^2 - 0.27 \left( \frac{20}{3} \right)^3 = 40 \text{ m}\).

Calculate the displacement at \(t = 10\):

\(s = 2.7(10)^2 - 0.27(10)^3 = 270 - 270 = 0 \text{ m}\).

The total distance travelled is the sum of the absolute displacements: \(40 + 40 = 80 \text{ m}\).

(i) To find the maximum velocity during the first 2 seconds, differentiate \(v = 12t - 4t^2\) with respect to \(t\):

\(\frac{dv}{dt} = 12 - 8t\).

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

\(12 - 8t = 0 \Rightarrow t = 1.5\).

Substitute \(t = 1.5\) into the velocity equation:

\(v = 12 \times 1.5 - 4 \times (1.5)^2 = 9\) m s^-1.

(ii) For \(0 \leq t \leq 2\), the acceleration is \(\frac{dv}{dt} = 12 - 8t\). At \(t = 2\), \(\frac{dv}{dt} = 12 - 8 \times 2 = -4\).

For \(2 \leq t \leq 4\), the acceleration is \(\frac{dv}{dt} = -4\).

Since the acceleration is \(-4\) in both cases, there is no instantaneous change at \(t = 2\).

(iii) The velocity-time graph is a quadratic curve for \(0 \leq t \leq 2\) with a maximum at \(t = 1.5\), and a straight line with a negative gradient from \(t = 2\) to \(t = 4\).

(iv) The distance travelled is the area under the velocity-time graph. For \(0 \leq t \leq 2\), integrate \(12t - 4t^2\):

\(\int (12t - 4t^2) \, dt = 6t^2 - \frac{4}{3}t^3\) from 0 to 2.

\(= 6 \times 2^2 - \frac{4}{3} \times 2^3 = 16\).

For \(2 \leq t \leq 4\), the area of the triangle is \(\frac{1}{2} \times 2 \times 8 = 8\).

\(Total distance = 16 + 8 = 24.\)

Distance travelled is \(\frac{64}{3}\) m or 21.3 m.

(i) The velocity function for \(0 \leq t \leq 10\) is \(v = 4 + 0.2t\). The acceleration is the derivative of velocity with respect to time:

\(a = \frac{dv}{dt} = 0.2\) m s^-2.

(ii) For \(10 \leq t \leq 20\), the velocity function is \(v = -2 + \frac{800}{t^2}\). The acceleration is:

\(a = \frac{dv}{dt} = -\frac{d}{dt}\left(2 - \frac{800}{t^2}\right) = \frac{1600}{t^3}\).

At \(t = 20\), \(a = \frac{1600}{20^3} = -0.2\) m s^-2.

(iii) The velocity-time graph consists of a straight line from \((0, 4)\) to \((10, 6)\) and a curve from \((10, 6)\) to \((20, 0)\) with correct concavity.

(iv) The total distance is the area under the velocity-time graph. For \(0 \leq t \leq 10\), the area is a trapezium:

\(\text{Area} = \frac{1}{2} \times (4 + 6) \times 10 = 50\) m.

For \(10 \leq t \leq 20\), integrate \(v = -2 + \frac{800}{t^2}\):

\(\int_{10}^{20} \left(-2 + \frac{800}{t^2}\right) dt = \left[-2t - \frac{800}{t}\right]_{10}^{20} = (-40 - 40 + 20 + 80) = 20\) m.

\(Total distance = 50 + 20 = 70 m.\)