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)`

Explanation

To solve the problem step by step, we will first identify the given data and then apply Newton's second law to find the acceleration, and subsequently the velocity vector after 6 seconds.

### Step 1: Identify the given data
- Mass of the object, \( m = 0.2 \, \text{kg} \)
- Force applied, \( \vec{F} = 0.3 \hat{i} - 0.4 \hat{j} \, \text{N} \)
- Time, \( t = 6 \, \text{s} \)

### Step 2: Calculate the acceleration
Using Newton's second law, we can find the acceleration \( \vec{a} \) using the formula:
\[ \vec{a} = \frac{\vec{F}}{m} \]

Calculating the components of acceleration:
- The x-component of the force is \( F_x = 0.3 \, \text{N} \)
- The y-component of the force is \( F_y = -0.4 \, \text{N} \)

Now, calculate \( a_x \) and \( a_y \):
\[ a_x = \frac{F_x}{m} = \frac{0.3}{0.2} = 1.5 \, \text{m/s}^2 \]
\[ a_y = \frac{F_y}{m} = \frac{-0.4}{0.2} = -2 \, \text{m/s}^2 \]

### Step 3: Calculate the velocity components after 6 seconds
Since the object starts from rest, the initial velocity \( \vec{u} = 0 \, \text{m/s} \). We can use the equation of motion:
\[ \vec{v} = \vec{u} + \vec{a} \cdot t \]

Calculating the x-component of velocity \( v_x \):
\[ v_x = u_x + a_x \cdot t = 0 + (1.5) \cdot (6) = 9 \, \text{m/s} \]

Calculating the y-component of velocity \( v_y \):
\[ v_y = u_y + a_y \cdot t = 0 + (-2) \cdot (6) = -12 \, \text{m/s} \]

### Step 4: Write the velocity vector
Combining the components, the velocity vector \( \vec{v} \) after 6 seconds is:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} = 9 \hat{i} - 12 \hat{j} \, \text{m/s} \]

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

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