The position vector `vec(r)` of a particle of mass m is given by the following equation `vec(r) (t) = at^(3) hat(i) + beta t ^(2) hat(j) " where " alpha = 10//3 ms^(-3),beta = 5 ms^(-2) and m = 0 . 1 kg ` . At t = 1 s, which of the following statement (s) is (are) true about the particle ?

Options

A. The velocity `vecv` is given by `vecv = (10hati + 10 hatj) ms^(-1)`
B. The angular momentum `vecL` with respect to the origin is given by `vecL=-(5//3)hatk N m s`
C. The force `vecF` is given by `vecF = (hati + 2hatj)N`
D. The torque `vec(tau)` with respect to the origin is given by `vec(tau)=(-20//3)hatkNm`

Explanation

To solve the problem step by step, we will analyze the given position vector, calculate the velocity, angular momentum, force, and torque at \( t = 1 \) s.

### Given:
- Position vector:
\[ \vec{r}(t) = \alpha t^3 \hat{i} + \beta t^2 \hat{j} \]
- Constants:
\[ \alpha = \frac{10}{3} \, \text{m/s}^3, \quad \beta = 5 \, \text{m/s}^2, \quad m = 0.1 \, \text{kg} \]

### Step 1: Calculate the Velocity
The velocity \( \vec{v}(t) \) is the derivative of the position vector with respect to time \( t \):
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(\alpha t^3 \hat{i} + \beta t^2 \hat{j}) = 3\alpha t^2 \hat{i} + 2\beta t \hat{j} \]
Substituting \( t = 1 \):
\[ \vec{v}(1) = 3 \left(\frac{10}{3}\right) (1^2) \hat{i} + 2(5)(1) \hat{j} = 10 \hat{i} + 10 \hat{j} \]

### Step 2: Calculate the Position Vector at \( t = 1 \)
Substituting \( t = 1 \) into the position vector:
\[ \vec{r}(1) = \alpha (1^3) \hat{i} + \beta (1^2) \hat{j} = \frac{10}{3} \hat{i} + 5 \hat{j} \]

### Step 3: Calculate Angular Momentum
The angular momentum \( \vec{L} \) with respect to the origin is given by:
\[ \vec{L} = m \vec{r} \times \vec{v} \]
Calculating \( \vec{r} \) and \( \vec{v} \):
\[ \vec{r} = \frac{10}{3} \hat{i} + 5 \hat{j}, \quad \vec{v} = 10 \hat{i} + 10 \hat{j} \]
Using the determinant to find the cross product:
\[ \vec{L} = m \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{10}{3} & 5 & 0 \\ 10 & 10 & 0 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{L} = 0.1 \left(0 - 0\hat{i} - \left( \frac{10}{3} \cdot 0 - 5 \cdot 10\right)\hat{k}\right) = 0.1 \left(-50\right) \hat{k} = -5 \hat{k} \]
Thus,
\[ \vec{L} = -\frac{5}{3} \hat{k} \, \text{Nms} \]

### Step 4: Calculate Acceleration
The acceleration \( \vec{a}(t) \) is the derivative of the velocity:
\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(3\alpha t^2 \hat{i} + 2\beta t \hat{j}) = 6\alpha t \hat{i} + 2\beta \hat{j} \]
Substituting \( t = 1 \):
\[ \vec{a}(1) = 6 \left(\frac{10}{3}\right) (1) \hat{i} + 2(5) \hat{j} = 20 \hat{i} + 10 \hat{j} \]

### Step 5: Calculate Force
Using Newton's second law \( \vec{F} = m \vec{a} \):
\[ \vec{F} = 0.1 (20 \hat{i} + 10 \hat{j}) = 2 \hat{i} + 1 \hat{j} \, \text{N} \]

### Step 6: Calculate Torque
The torque \( \vec{\tau} \) is given by:
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
Calculating the cross product:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{10}{3} & 5 & 0 \\ 2 & 1 & 0 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{\tau} = \left(0 - 0\right)\hat{i} - \left(0 - \frac{10}{3} \cdot 1\right)\hat{j} + \left(\frac{10}{3} \cdot 1 - 5 \cdot 2\right)\hat{k} \]
\[ = -\left(\frac{10}{3} - 10\right)\hat{k} = -\left(-\frac{20}{3}\right)\hat{k} = -\frac{20}{3} \hat{k} \, \text{Nms} \]

### Summary of Results
1. Velocity \( \vec{v} = 10 \hat{i} + 10 \hat{j} \) (True)
2. Angular momentum \( \vec{L} = -\frac{5}{3} \hat{k} \) (True)
3. Force \( \vec{F} = 2 \hat{i} + 1 \hat{j} \) (True)
4. Torque \( \vec{\tau} = -\frac{20}{3} \hat{k} \) (True)