An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin. When the object is at `x=+2m`,its velocity is `-4hat j m/s`.The object's velocity (v) and acceleration (a) at `x=-2m` will be

Options

A. `v=-4jm//s,a=-8hat jm//s^(2)`
B. `v=4hat i m//s,a=8hat j m//s^(2)`
C. `v=4hat j m//s,a=8hat i m//s^(2)`
D. `v=-4jm//s,a=8im//s^(2)`

Answer

To solve the problem, we need to analyze the motion of the object moving in a circular path. Let's break down the steps:

### Step 1: Understand the motion
The object is moving in a circular path with a constant speed. This means that while the speed remains constant, the direction of the velocity vector is continuously changing.

### Step 2: Identify the given information
- The object is at position \( x = +2 \, \text{m} \) with a velocity \( \vec{v} = -4 \hat{j} \, \text{m/s} \).
- This indicates that when the object is at \( x = +2 \, \text{m} \), it is moving downward (in the negative y-direction).

### Step 3: Determine the radius of the circular path
Since the object is moving in a circular path centered at the origin, the radius \( r \) can be determined from the position \( x = +2 \, \text{m} \). The radius is \( r = 2 \, \text{m} \).

### Step 4: Find the position at \( x = -2 \, \text{m} \)
When the object is at \( x = -2 \, \text{m} \), it will be on the opposite side of the circle. The coordinates at this position will be \( (-2, 0) \).

### Step 5: Determine the velocity at \( x = -2 \, \text{m} \)
Since the object is moving with constant speed, we need to find the direction of the velocity vector at this position. The velocity vector will be tangential to the circular path.

At \( x = +2 \, \text{m} \), the velocity vector is downward, which means the object is moving counterclockwise. Therefore, at \( x = -2 \, \text{m} \), the object will be moving to the right (in the positive x-direction).

Thus, the velocity at \( x = -2 \, \text{m} \) will be:
\[ \vec{v} = 4 \hat{i} \, \text{m/s} \]

### Step 6: Determine the acceleration at \( x = -2 \, \text{m} \)
The acceleration of an object moving in a circular path is directed towards the center of the circle (centripetal acceleration). The magnitude of the centripetal acceleration \( a \) can be calculated using the formula:
\[ a = \frac{v^2}{r} \]
Where \( v = 4 \, \text{m/s} \) and \( r = 2 \, \text{m} \).

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

The direction of the acceleration will be towards the center of the circular path, which is at the origin. Therefore, at \( x = -2 \, \text{m} \), the acceleration vector will be directed towards the origin, which can be represented as:
\[ \vec{a} = -4 \hat{i} \, \text{m/s}^2 \]

### Final Answer
- Velocity at \( x = -2 \, \text{m} \): \( \vec{v} = 4 \hat{i} \, \text{m/s} \)
- Acceleration at \( x = -2 \, \text{m} \): \( \vec{a} = -4 \hat{i} \, \text{m/s}^2 \)