Given \(x = 8 + 12t - t^3\). Velocity \(v = \frac{dx}{dt} = 12 - 3t^2\). Acceleration \(a = \frac{dv}{dt} = -6t\). When (v=0), \(12 - 3t^2 = 0 \Rightarrow t^2 = 4 \Rightarrow t = 2 \text{ s}\). At \(t=2 \text{ s}\), \(a = -6(2) = -12 \text{ m/s}^2\). Retardation is \(-a = 12 \text{ m/s}^2\).

Given \(x = (t + 5)^{-1}\). Velocity \(v = \frac{dx}{dt} = -(t + 5)^{-2}\). Acceleration \(a = \frac{dv}{dt} = 2(t + 5)^{-3}\). From \(v = -(t + 5)^{-2}\), we have \((t+5)^{-1} = ((-v)^{-1/2})\). So \(a = 2((t+5)^{-1})^3 = 2((-v)^{-1/2})^3 = 2(-v)^{3/2}\). Thus, \(a \propto (\text{Velocity})^{3/2}\).

Given \(x = 5t - 2t^2\) and \(y = 10t\). Differentiating twice with respect to time to find acceleration. \(v_x = \frac{dx}{dt} = 5 - 4t\), \(a_x = \frac{dv_x}{dt} = -4\text{ m/s}^2\). \(v_y = \frac{dy}{dt} = 10\), \(a_y = \frac{dv_y}{dt} = 0\text{ m/s}^2\). The acceleration vector is \(\vec{a} = -4\hat{i}\). The x-component of acceleration is \(-4\text{ m/s}^2\), which is constant.

Velocity \(\vec{v}(t) = \frac{d\vec{r}}{dt} = 4\hat{i} + 4\that{j}\). At \(t = 1 \text{s}\), \(\vec{v} = 4\hat{i} + 4\hat{j}\). Magnitude \(v = \sqrt{4^2 + 4^2} = 4\sqrt{2} \text{m/s}\). The angle with the x-axis is \(tan\theta = \frac{v_y}{v_x} = \frac{4}{4} = 1 ⇒
\theta = 45^\circ\).

The velocity components are \(v_x = \frac{dx}{dt} = 6t\) and \(v_y = \frac{dy}{dt} = 5\). At \(t = 2\) s, \(v_x = 12\text{ m/s}\) and \(v_y = 5\text{ m/s}\). Speed is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 5^2} = 13\text{ m/s}\).