A `0.2 kg` object at rest is subjected to a force `(0.3hati-0.4hatj) N`. What is its velocity vector after `6 sec`

Options

A. `(9hati-12hatj)`
B. `(8hati-16hatj)`
C. `(12hati-9hatj)`
D. `(16hati-8hatj)`

Answer

To solve the problem, we need to determine the velocity vector of a 0.2 kg object after being subjected to a force of \( (0.3 \hat{i} - 0.4 \hat{j}) \) N for 6 seconds. The object starts from rest.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Mass of the object, \( m = 0.2 \, \text{kg} \)
- Force applied, \( \vec{F} = (0.3 \hat{i} - 0.4 \hat{j}) \, \text{N} \)
- Time duration, \( t = 6 \, \text{s} \)
- Initial velocity, \( \vec{u} = 0 \, \text{m/s} \) (since the object is at rest)

2. **Calculate the Acceleration:**
- Using Newton's second law, \( \vec{F} = m \vec{a} \), we can find the acceleration \( \vec{a} \):
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{(0.3 \hat{i} - 0.4 \hat{j})}{0.2} \]
- Calculate the components:
\[ a_x = \frac{0.3}{0.2} = 1.5 \, \text{m/s}^2 \]
\[ a_y = \frac{-0.4}{0.2} = -2 \, \text{m/s}^2 \]
- Therefore, the acceleration vector is:
\[ \vec{a} = (1.5 \hat{i} - 2 \hat{j}) \, \text{m/s}^2 \]

3. **Calculate the Final Velocity:**
- Using the equation of motion \( \vec{v} = \vec{u} + \vec{a} t \):
\[ \vec{v} = 0 + (1.5 \hat{i} - 2 \hat{j}) \cdot 6 \]
- Calculate the components:
\[ v_x = 1.5 \cdot 6 = 9 \, \text{m/s} \]
\[ v_y = -2 \cdot 6 = -12 \, \text{m/s} \]
- Therefore, the final velocity vector is:
\[ \vec{v} = 9 \hat{i} - 12 \hat{j} \, \text{m/s} \]

4. **Final Result:**
- The velocity vector after 6 seconds is:
\[ \vec{v} = 9 \hat{i} - 12 \hat{j} \, \text{m/s} \]