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. `18hat(i)+3hat(j)`
B. `18hat(i)+6hat(j)`
C. `3hat(i)+18hat(j)`
D. `19hat(i)+4hat(j)`

Answer

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

### Step 1: Identify the given values
- 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 acceleration
Using Newton's second law, the acceleration \( \vec{a} \) can be calculated using the formula:
\[ \vec{a} = \frac{\vec{F}}{m} \]
Substituting the values:
\[ \vec{a} = \frac{6t^2 \hat{i} + 4t \hat{j}}{3} \]
This simplifies to:
\[ \vec{a} = 2t^2 \hat{i} + \frac{4}{3}t \hat{j} \]

### Step 3: Relate acceleration to velocity
Acceleration can also be expressed as:
\[ \vec{a} = \frac{d\vec{v}}{dt} \]
Thus, we can write:
\[ d\vec{v} = \vec{a} \, dt \]
Substituting for \( \vec{a} \):
\[ d\vec{v} = \left(2t^2 \hat{i} + \frac{4}{3}t \hat{j}\right) dt \]

### Step 4: Integrate to find velocity
Integrating both sides from \( t = 0 \) to \( t = 3 \) seconds:
\[ \int_{0}^{V} d\vec{v} = \int_{0}^{3} \left(2t^2 \hat{i} + \frac{4}{3}t \hat{j}\right) dt \]
This gives:
\[ \vec{v} = \left[ \frac{2}{3}t^3 \hat{i} + \frac{4}{6}t^2 \hat{j} \right]_{0}^{3} \]

### Step 5: Evaluate the integral
Calculating the limits:
\[ \vec{v} = \left( \frac{2}{3}(3^3) \hat{i} + \frac{4}{6}(3^2) \hat{j} \right) - \left( \frac{2}{3}(0^3) \hat{i} + \frac{4}{6}(0^2) \hat{j} \right) \]
This simplifies to:
\[ \vec{v} = \left( \frac{2}{3}(27) \hat{i} + \frac{4}{6}(9) \hat{j} \right) \]
\[ \vec{v} = 18 \hat{i} + 6 \hat{j} \]

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