The initial position of an object at rest is given by `3 hati - 8 hatj`. It moves with constant acceleration and reaches to the position `2 hati +4hatj` after 4s. What is its acceleration ?

Options

A. `-(1)/(8) hati + (3)/(2) hatj`
B. `2 hati - (1)/(8) hatj`
C. `-(1)/(2) hati + 8 hatj`
D. `8 hati -(3)/(2) hatj`

Explanation

To find the acceleration of the object, we can use the equations of motion. Let's break it down step by step.

### Step 1: Identify the Initial and Final Positions
The initial position of the object is given as:
\[ \mathbf{r_1} = 3 \hat{i} - 8 \hat{j} \]
The final position after 4 seconds is:
\[ \mathbf{r_2} = 2 \hat{i} + 4 \hat{j} \]

### Step 2: Calculate the Change in Position
The change in position vector \(\Delta \mathbf{r}\) can be calculated as:
\[ \Delta \mathbf{r} = \mathbf{r_2} - \mathbf{r_1} \]
Substituting the values:
\[ \Delta \mathbf{r} = (2 \hat{i} + 4 \hat{j}) - (3 \hat{i} - 8 \hat{j}) \]
\[ = 2 \hat{i} + 4 \hat{j} - 3 \hat{i} + 8 \hat{j} \]
\[ = -1 \hat{i} + 12 \hat{j} \]

### Step 3: Use the Second Equation of Motion
The second equation of motion states:
\[ \Delta \mathbf{r} = \mathbf{u} t + \frac{1}{2} \mathbf{a} t^2 \]
Where:
- \(\Delta \mathbf{r} = -1 \hat{i} + 12 \hat{j}\)
- \(\mathbf{u} = 0\) (initial velocity, since the object is at rest)
- \(t = 4 \, \text{s}\)

Substituting the known values into the equation:
\[ -1 \hat{i} + 12 \hat{j} = 0 + \frac{1}{2} \mathbf{a} (4^2) \]
\[ -1 \hat{i} + 12 \hat{j} = \frac{1}{2} \mathbf{a} (16) \]
\[ -1 \hat{i} + 12 \hat{j} = 8 \mathbf{a} \]

### Step 4: Solve for Acceleration
Now, we can solve for \(\mathbf{a}\):
\[ \mathbf{a} = \frac{-1 \hat{i} + 12 \hat{j}}{8} \]
Breaking it down:
\[ \mathbf{a} = \frac{-1}{8} \hat{i} + \frac{12}{8} \hat{j} \]
\[ \mathbf{a} = -\frac{1}{8} \hat{i} + \frac{3}{2} \hat{j} \]

### Conclusion
The acceleration of the object is:
\[ \mathbf{a} = -\frac{1}{8} \hat{i} + \frac{3}{2} \hat{j} \]