The force on a body of mass 1 kg is `(20 hati + 10hatj)` N. If is starts from rest, then the position of the body at time t = 2 s, is

Options

A. `-20 hati- 40 hatj`
B. `20 hati - 40 hatj`
C. `40 hati - 20 hatj`
D. `40 hati + 20 hatj`

Explanation

To find the position of the body at time \( t = 2 \) seconds, we will follow these steps:

### Step 1: Identify the Given Information
We have:
- Mass of the body, \( m = 1 \, \text{kg} \)
- Force acting on the body, \( \vec{F} = 20 \hat{i} + 10 \hat{j} \, \text{N} \)

### Step 2: Calculate Acceleration
Using Newton's second law, \( \vec{F} = m \vec{a} \), we can find the acceleration \( \vec{a} \):
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{20 \hat{i} + 10 \hat{j}}{1} = 20 \hat{i} + 10 \hat{j} \, \text{m/s}^2 \]

### Step 3: Use the Equations of Motion
Since the body starts from rest, the initial velocity \( \vec{u} = 0 \). We will use the equation of motion to find the position after \( t = 2 \) seconds:
\[ \vec{s} = \vec{u} t + \frac{1}{2} \vec{a} t^2 \]
Substituting the values:
\[ \vec{s} = 0 + \frac{1}{2} (20 \hat{i} + 10 \hat{j}) (2^2) \]

### Step 4: Calculate the Position Vector
Calculating \( t^2 = 4 \):
\[ \vec{s} = \frac{1}{2} (20 \hat{i} + 10 \hat{j}) \cdot 4 = (10 \hat{i} + 5 \hat{j}) \cdot 4 = 40 \hat{i} + 20 \hat{j} \]

### Step 5: Final Position
Thus, the position of the body at \( t = 2 \) seconds is:
\[ \vec{s} = 40 \hat{i} + 20 \hat{j} \, \text{m} \]

### Conclusion
The final position vector is \( 40 \hat{i} + 20 \hat{j} \).

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