(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 \(t = 2.5\).