An object moves at a constant speed along a circular path in a horizontal `XY` plane, with the center at the origin. When the object is at `x=-2m`, its velocity is `-(4m//s)hatj`. What is the object's acceleration when it is `y=2m`

Options

A. `-(8m//s^(2))vecj`
B. `-(8m//s^(2))veci`
C. `-(4m//s^(2))vecj`
D. `-(4m//s^(2))veci`

Explanation

To solve the problem step by step, we need to analyze the motion of the object moving in a circular path and determine its acceleration when it is at a specific position.

### Step 1: Understand the motion
The object is moving in a circular path with a constant speed. The center of the circular path is at the origin (0,0) in the XY plane.

### Step 2: Identify the given information
- When the object is at `x = -2 m`, its velocity is `-4 m/s` in the `-j` direction (downward).
- We need to find the acceleration when the object is at `y = 2 m`.

### Step 3: Determine the radius of the circular path
Since the object is at `x = -2 m`, we can find the radius of the circular path. The distance from the center (origin) to this point is:
\[ r = \sqrt{(-2)^2 + 0^2} = 2 \text{ m} \]

### Step 4: Calculate the speed
The speed of the object is given as `4 m/s`. This speed remains constant as the object moves along the circular path.

### Step 5: Use the formula for centripetal acceleration
The formula for centripetal acceleration \( a \) is given by:
\[ a = \frac{v^2}{r} \]
where:
- \( v \) is the speed of the object,
- \( r \) is the radius of the circular path.

### Step 6: Substitute the known values
Substituting the values we have:
- \( v = 4 \text{ m/s} \)
- \( r = 2 \text{ m} \)

Calculating the acceleration:
\[ a = \frac{(4 \text{ m/s})^2}{2 \text{ m}} = \frac{16 \text{ m}^2/\text{s}^2}{2 \text{ m}} = 8 \text{ m/s}^2 \]

### Step 7: Determine the direction of acceleration
In circular motion, the acceleration is directed towards the center of the circular path. Since the object is at `y = 2 m`, the center is at the origin `(0,0)`, which means the acceleration vector points in the negative `j` direction (downward).

### Step 8: Write the final answer
Thus, the acceleration of the object when it is at `y = 2 m` is:
\[ \mathbf{a} = -8 \hat{j} \text{ m/s}^2 \]

### Final Answer:
The object's acceleration when it is at `y = 2 m` is:
\[ \mathbf{a} = -8 \hat{j} \text{ m/s}^2 \]

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