A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in moduli. At the initial moment t=0 the speed of the particle equals `V_(0)` then
The total acceleration of the particle as function of velocity and distance covered

Options

A. `a= sqrt(2)(v^(2))/(R)`
B. `a= sqrt(2)(v)/(R)`
C. `a= sqrt(2)(R)/(v)`
D. `a= (2R)/(v)`

Answer

To solve the problem, we need to derive the total acceleration of a particle moving in a circular path with deceleration, where the tangential and normal (centripetal) accelerations are equal in magnitude.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The particle moves in a circular path of radius \( R \).
- At any moment, the tangential acceleration \( a_t \) and the normal (centripetal) acceleration \( a_n \) are equal in magnitude.
- At time \( t = 0 \), the speed of the particle is \( V_0 \).

2. **Expressing the Accelerations**:
- The tangential acceleration \( a_t \) is related to the change in speed of the particle, and since the particle is decelerating, we can denote it as:
\[ a_t = -k \quad \text{(where \( k \) is a positive constant)} \]
- The normal (centripetal) acceleration \( a_n \) is given by:
\[ a_n = \frac{v^2}{R} \]
- Given that \( |a_t| = |a_n| \), we can equate the two:
\[ k = \frac{v^2}{R} \]

3. **Finding the Relationship Between Speed and Time**:
- Since the tangential acceleration is negative (deceleration), we can write:
\[ \frac{dv}{dt} = -k \]
- Integrating this with respect to time gives:
\[ v = V_0 - kt \]

4. **Finding the Distance Covered**:
- The distance \( s \) covered can be expressed as:
\[ s = \int v \, dt = \int (V_0 - kt) \, dt = V_0 t - \frac{1}{2} kt^2 \]

5. **Substituting for \( k \)**:
- From \( k = \frac{v^2}{R} \), we can express \( k \) in terms of \( v \):
\[ k = \frac{(V_0 - kt)^2}{R} \]
- Rearranging gives us a relationship between \( v \), \( R \), and \( s \).

6. **Total Acceleration**:
- The total acceleration \( a \) of the particle can be expressed as:
\[ a = \sqrt{a_t^2 + a_n^2} \]
- Substituting the expressions for \( a_t \) and \( a_n \):
\[ a = \sqrt{(-k)^2 + \left(\frac{v^2}{R}\right)^2} \]
- Since \( k = \frac{v^2}{R} \), we have:
\[ a = \sqrt{\left(\frac{v^2}{R}\right)^2 + \left(\frac{v^2}{R}\right)^2} = \sqrt{2 \left(\frac{v^2}{R}\right)^2} = \frac{v^2 \sqrt{2}}{R} \]

7. **Final Expression**:
- Thus, the total acceleration \( a \) as a function of velocity \( v \) and distance covered \( s \) is:
\[ a = \frac{v^2 \sqrt{2}}{R} \]