(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