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

Options

A. `-34 (hat(k) -hat(i))`
B. `-48hat(k)`
C. `36hat(k)`
D. `48(hat(i) + hat(j))`

Explanation

To find the angular momentum of a particle with respect to the origin, we can follow these steps:

### Step 1: Write down the position vector
The position vector of the particle is given by:
\[ \vec{r}(t) = 2t \hat{i} - 3t^2 \hat{j} \]

### Step 2: Find the velocity vector
The velocity vector \(\vec{v}(t)\) is the derivative of the position vector with respect to time \(t\):
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(2t \hat{i} - 3t^2 \hat{j}) \]
Calculating the derivative:
\[ \vec{v}(t) = 2 \hat{i} - 6t \hat{j} \]

### Step 3: Evaluate the position and velocity at \(t = 2\)
Now we substitute \(t = 2\) into both the position and velocity vectors:
\[ \vec{r}(2) = 2(2) \hat{i} - 3(2^2) \hat{j} = 4 \hat{i} - 12 \hat{j} \]
\[ \vec{v}(2) = 2 \hat{i} - 6(2) \hat{j} = 2 \hat{i} - 12 \hat{j} \]

### Step 4: Calculate the angular momentum
The angular momentum \(\vec{L}\) with respect to the origin is given by:
\[ \vec{L} = m \vec{r} \times \vec{v} \]
where \(m = 2 \, \text{kg}\).

First, we need to calculate the cross product \(\vec{r}(2) \times \vec{v}(2)\):
\[ \vec{r}(2) = 4 \hat{i} - 12 \hat{j}, \quad \vec{v}(2) = 2 \hat{i} - 12 \hat{j} \]
Now, we compute the cross product:
\[ \vec{r} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -12 & 0 \\ 2 & -12 & 0 \end{vmatrix} \]

### Step 5: Calculate the determinant
Calculating the determinant:
\[ \vec{r} \times \vec{v} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(4 \cdot (-12) - (-12) \cdot 2) \]
\[ = \hat{k}(-48 + 24) = -24 \hat{k} \]

### Step 6: Multiply by mass
Now, we multiply by the mass \(m\):
\[ \vec{L} = 2 \cdot (-24 \hat{k}) = -48 \hat{k} \]

### Final Answer
Thus, the angular momentum of the particle with respect to the origin at time \(t = 2\) is:
\[ \vec{L} = -48 \hat{k} \]

---