A particle moves along a horizontal circle with constant speed. If 'a' is its acceleration and 'E' is its kinetic energy
A) a is constant B) E is constant C) a is variable D) E is variable

Options

A. A & B are correct
B. C & D are correct
C. A & D are correct
D. B & C are correct

Answer

To solve the problem, we need to analyze the motion of a particle moving in a horizontal circle with constant speed. We'll look at the acceleration and kinetic energy of the particle.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is moving in a horizontal circle at a constant speed. This means that while the speed (magnitude of velocity) remains the same, the direction of the velocity is continuously changing.

**Hint**: Remember that speed is a scalar quantity, while velocity is a vector quantity that includes both magnitude and direction.

2. **Analyzing Velocity**:
- Since the particle is moving in a circle, its velocity is changing due to the change in direction. Even though the speed is constant, the velocity vector is not constant because its direction is changing.

**Hint**: Consider how velocity changes when an object moves in a circular path.

3. **Determining Acceleration**:
- Acceleration is defined as the rate of change of velocity. Since the velocity is changing (due to the changing direction), the particle experiences centripetal acceleration directed towards the center of the circle. This acceleration is constant in magnitude but changes direction as the particle moves around the circle.

**Hint**: Recall that centripetal acceleration is given by \( a = \frac{v^2}{r} \), where \( v \) is the constant speed and \( r \) is the radius of the circle.

4. **Kinetic Energy Calculation**:
- The kinetic energy \( E \) of the particle is given by the formula \( E = \frac{1}{2} mv^2 \). Since the speed \( v \) is constant, the kinetic energy remains constant as well.

**Hint**: Remember that kinetic energy depends only on the magnitude of velocity, not its direction.

5. **Conclusion**:
- From the analysis, we find that:
- The acceleration \( a \) is variable in direction (but constant in magnitude).
- The kinetic energy \( E \) is constant since the speed is constant.

Therefore, the correct options are:
- A is variable (acceleration is variable in direction).
- B is constant (kinetic energy is constant).

### Final Answer:
The correct option is **B and C are correct**.