A particle of mass 'm' is moving along a circle of radius 'r'. At some instant, its speed is 'v' and it is gaining speed at a uniform rate'a', then, at the given instant, acceleration of the particle is :

Options

A. along the radius
B. inclined to radius at `theta="sin"^(-1)(1)/([1+(v^(4))/(a^(2)r^(2))]^(1//2))`
C. inclined to radius at `theta=cos^(-1)(ar)/(v^(2))`
D. inclined to radius at `theta=tan^(-1)(v^(2))/(ar)`

Answer

To find the acceleration of a particle moving in a circular path with a given speed and a uniform rate of acceleration, we can break down the problem into steps.

### Step-by-Step Solution:

1. **Identify the Components of Acceleration**:
- The particle has two components of acceleration:
- **Centripetal Acceleration (a_c)**: This is due to the circular motion and is directed towards the center of the circle. It can be calculated using the formula:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the speed of the particle and \( r \) is the radius of the circle.

2. **Determine Tangential Acceleration (a_t)**:
- The particle is gaining speed at a uniform rate \( a \). This tangential acceleration is given as:
\[ a_t = a \]
where \( a \) is the uniform rate of acceleration.

3. **Calculate the Net Acceleration (a_net)**:
- The total or net acceleration of the particle can be found using the Pythagorean theorem, since the tangential and centripetal accelerations are perpendicular to each other:
\[ a_{net} = \sqrt{a_t^2 + a_c^2} \]
- Substituting the values of \( a_t \) and \( a_c \):
\[ a_{net} = \sqrt{a^2 + \left(\frac{v^2}{r}\right)^2} \]

4. **Final Expression**:
- Therefore, the expression for the acceleration of the particle at the given instant is:
\[ a_{net} = \sqrt{a^2 + \frac{v^4}{r^2}} \]

### Final Answer:
The acceleration of the particle is:
\[ a_{net} = \sqrt{a^2 + \frac{v^4}{r^2}} \]