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. ` 18 hat I + 3 hat j`
B. ` 18 hat I + 6 hat j`
C. 3 hat I + 1 8 hat j`
D. ` 18 hat I + 4har j`

Explanation

To find the velocity of the object at \( t = 3 \) seconds, we can follow these steps:

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

### Step 2: Calculate the acceleration
Using Newton's second law, \( \vec{F} = m \vec{a} \), we can find the acceleration \( \vec{a} \):
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{6t^2 \hat{i} + 4t \hat{j}}{3} \]
This simplifies to:
\[ \vec{a} = 2t^2 \hat{i} + \frac{4}{3}t \hat{j} \]

### Step 3: Relate acceleration to velocity
Acceleration is the derivative of velocity with respect to time:
\[ \vec{a} = \frac{d\vec{v}}{dt} \]
Thus, we can write:
\[ \frac{d\vec{v}}{dt} = 2t^2 \hat{i} + \frac{4}{3}t \hat{j} \]

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

### Step 5: Calculate velocity at \( t = 3 \, \text{s} \)
Substituting \( t = 3 \):
\[ \vec{v} = \frac{2}{3}(3^3) \hat{i} + \frac{2}{3}(3^2) \hat{j} \]
Calculating each term:
\[ \vec{v} = \frac{2}{3}(27) \hat{i} + \frac{2}{3}(9) \hat{j} \]
\[ \vec{v} = 18 \hat{i} + 6 \hat{j} \]

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

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