An object of mass ` 3 kg ` is at rest. Now a force of ` vec F = 6 t^2 hat I + 4 t hat j` is applied on the object, the velocity of object at `t= 3 s` is.

Options

A. `18hatj+3hatj`
B. `18hati+6hatj`
C. `3hati+18hatj`
D. `18hati+4hatj`

Explanation

To solve the problem step by step, we will follow the principles of Newton's laws of motion and calculus.

### Step 1: Identify the given information
- Mass of the object, \( m = 3 \, \text{kg} \)
- Force applied, \( \vec{F} = 6t^2 \hat{i} + 4t \hat{j} \)
- Time, \( t = 3 \, \text{s} \)

### Step 2: Calculate the acceleration
Using Newton's second law, we know that:
\[ \vec{F} = m \vec{a} \]
Thus, the acceleration \( \vec{a} \) can be calculated as:
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{6t^2 \hat{i} + 4t \hat{j}}{3} \]
Simplifying this gives:
\[ \vec{a} = 2t^2 \hat{i} + \frac{4}{3}t \hat{j} \]

### Step 3: Integrate acceleration to find velocity
Since acceleration is the derivative of velocity with respect to time, we can write:
\[ \vec{a} = \frac{d\vec{v}}{dt} \]
Integrating both sides with respect to time \( t \):
\[ \vec{v} = \int \vec{a} \, dt = \int (2t^2 \hat{i} + \frac{4}{3}t \hat{j}) \, dt \]
This gives us:
\[ \vec{v} = \left(\frac{2}{3}t^3 \hat{i} + \frac{2}{3}t^2 \hat{j}\right) + \vec{C} \]
Where \( \vec{C} \) is the constant of integration. Since the object is at rest initially, \( \vec{v}(0) = 0 \), we find that \( \vec{C} = 0 \).

### Step 4: Substitute \( t = 3 \, \text{s} \) into the velocity equation
Now we substitute \( t = 3 \) into the velocity equation:
\[ \vec{v} = \left(\frac{2}{3}(3^3) \hat{i} + \frac{2}{3}(3^2) \hat{j}\right) \]
Calculating this gives:
\[ \vec{v} = \left(\frac{2}{3} \cdot 27 \hat{i} + \frac{2}{3} \cdot 9 \hat{j}\right) = (18 \hat{i} + 6 \hat{j}) \]

### Final Answer
Thus, the velocity of the object at \( t = 3 \, \text{s} \) is:
\[ \vec{v} = 18 \hat{i} + 6 \hat{j} \]