The time dependence of the position of a particle of mass m = 2 is given by $$\vec{r}(t) = 2t\hat{i} - 3t^2\hat{j}$$. Its angular momentum, with respect to the origin, at time t = 2 is:

We are given the position-time relation of the particle as $$\vec r(t)=2t\,\hat i-3t^{2}\,\hat j$$ and the mass of the particle as $$m=2\;\text{kg}$$. Our task is to calculate the angular momentum $$\vec L$$ of this particle about the origin at the instant $$t=2\ \text{s}$$.

First, recall the definition of angular momentum for a single particle about the origin:

$$\vec L=\vec r\times\vec p$$

where $$\vec r$$ is the position vector and $$\vec p$$ is the linear momentum. The linear momentum is related to velocity by $$\vec p=m\vec v$$. Therefore we must first obtain the velocity vector.

Velocity is the time derivative of position. Stating the formula:

$$\vec v(t)=\frac{d\vec r(t)}{dt}$$

Taking the derivative term-by-term, we have

$$\vec v(t)=\frac{d}{dt}\bigl(2t\,\hat i-3t^{2}\,\hat j\bigr) =2\,\hat i-6t\,\hat j.$$

Now we substitute $$t=2$$ to get the velocity at the required instant:

$$\vec v(2)=2\,\hat i-6(2)\,\hat j =2\,\hat i-12\,\hat j.$$

Next we calculate the linear momentum using $$\vec p=m\vec v$$:

$$\vec p(2)=m\vec v(2) =2\bigl(2\,\hat i-12\,\hat j\bigr) =4\,\hat i-24\,\hat j.$$

We also need the position vector at $$t=2$$. Substituting in the original expression, we get

$$\vec r(2)=2(2)\,\hat i-3(2)^{2}\,\hat j =4\,\hat i-12\,\hat j.$$

Now we compute the cross product $$\vec L=\vec r\times\vec p$$. Because both $$\vec r$$ and $$\vec p$$ lie in the $$xy$$-plane (their $$\hat k$$ components are zero), the cross product will point purely along $$\hat k$$. The standard determinant formula for the cross product in component form is

$$\vec r\times\vec p= \begin{vmatrix} \hat i & \hat j & \hat k\\ r_{x} & r_{y} & 0\\ p_{x} & p_{y} & 0 \end{vmatrix} =\bigl(r_{x}p_{y}-r_{y}p_{x}\bigr)\hat k.$$

Here $$r_{x}=4,\;r_{y}=-12,\;p_{x}=4,\;p_{y}=-24$$. Substituting these numerical values, we obtain

$$L_{z}=r_{x}p_{y}-r_{y}p_{x} =4(-24)-(-12)(4) =-96-(-48) =-96+48 =-48.$$

Therefore,

$$\vec L=-48\,\hat k.$$

Hence, the correct answer is Option C.