The volume of a spherical bubble is \(V = \frac{4}{3}\pi r^3\). Differentiating with respect to time gives \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\). Given \(r = 0.5t\), we have \(\frac{dr}{dt} = 0.5\), and at \(t = 4\text{ s}\), \(r = 2\). Thus, \(\frac{dV}{dt} = 4\pi (2)^2 (0.5) = 8\pi\text{ units/s}\).

Velocity \(v = 3t^2 - 12t + 3\) and acceleration \(a = 6t - 12\). Acceleration is zero at t = 2 s. Substituting \(t = 2\) in velocity equation gives \(v = 3(4) - 24 + 3 = -9 \text{ ms}^{-1}\).

Velocity is \(\vec{v} = \frac{d\vec{r}}{dt} = 4\hat{i} + 4t\hat{j}\). At \(t = 1\text{ s}\), \(\vec{v} = 4\hat{i} + 4\hat{j}\). Thus, magnitude \(v = \sqrt{4^2 + 4^2} = 4\sqrt{2}\text{ ms}^{-1}\) and angle \(\tan\theta = \frac{4}{4} = 1 ⇒ \theta = 45^\circ\) with respect to the x-axis.

Velocity is given by differentiating displacement with respect to time: \(v = \frac{ds}{dt} = 16t + 3\). For initial velocity, substitute \(t = 0\), which gives \(v = 3\text{ cm/s}\).

Velocity components: \(v_x = \frac{dx}{dt} = 7 + 8t\) and \(v_y = \frac{dy}{dt} = 5\). Acceleration components: \(a_x = \frac{dv_x}{dt} = 8\) and \(a_y = \frac{dv_y}{dt} = 0\). Total acceleration is \(a = \sqrt{a_x^2 + a_y^2} = 8\text{ m/s}^2\) at any time.

Integrating the equation \(\int_0^v \frac{dv}{A - Bv} = \int_0^t dt\) gives \(-\frac{1}{B} \ln(\frac{A - Bv}{A}) = t\). Simplifying for velocity gives \(v = \frac{A}{B}(1 - e^{-Bt})\).

Given \(v_x = \frac{dx}{dt} = c\) (constant), hence \(a_x = 0\). Differentiating \(y = ax^2\) gives \(v_y = 2ax\frac{dx}{dt} = 2axc\). Differentiating again, \(a_y = 2ac\frac{dx}{dt} = 2ac^2\). Therefore, total acceleration vector is \(a = 2ac^2\hat{j}\).

The particle comes to rest when \(v = 0\), giving \(4t - 2 = 0 ⇒ t = 0.5\text{ s}\). Integrating velocity to find position, \(x = int_0^t (4t - 2)dt = 2t^2 - 2t\). At \(t = 0.5\text{ s}\), \(x = 2(0.5)^2 - 2(0.5) = -0.5\text{ m}\).

Displacement is \(s = \int v \, dt = 6t - 1.5t^2\). To return to the starting point, \(s = 0 \implies 6t - 1.5t^2 = 0 \implies t = 4\text{ s}\).

Using \( a = v \frac{dv}{dx} \), we write \( v \frac{dv}{dx} = 5x \). Integrating both sides, \( \int v \, dv = \int 5x \, dx ⇒ \frac{v^2}{2} = \frac{5x^2}{2} + C \). Assuming the particle starts from rest, \( v^2 \propto x^2 ⇒ v \propto x^1 \).