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 th speed of the particle as a function of the distance covered S will be

Options

A. `v=v_(0) e^(-S//R)`
B. `v=v_(0)e^(S//R)`
C. `v=v_(0) e^(-R//S)`
D. `v=v_(0) e^(R//S)`

Explanation

To solve the problem, we need to find the speed of a particle moving in a circular path as a function of the distance covered, given that its tangential and normal accelerations are equal in magnitude.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The particle moves in a circular path with radius \( R \).
- At any moment, the tangential acceleration \( a_t \) and the normal (centripetal) acceleration \( a_n \) are equal in magnitude.
- The initial speed of the particle at \( t = 0 \) is \( v_0 \).

2. **Define the Accelerations**:
- The tangential acceleration \( a_t \) is given by:
\[ a_t = \frac{dv}{dt} \]
- The normal acceleration \( a_n \) (centripetal acceleration) is given by:
\[ a_n = \frac{v^2}{R} \]

3. **Setting Up the Equation**:
- Since the magnitudes of the tangential and normal accelerations are equal, we can write:
\[ \frac{dv}{dt} = \frac{v^2}{R} \]

4. **Changing Variables**:
- We want to express \( v \) as a function of the distance \( s \) covered by the particle. We can use the chain rule:
\[ \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} \]
- Here, \( \frac{ds}{dt} = v \), hence:
\[ \frac{dv}{dt} = \frac{dv}{ds} \cdot v \]

5. **Substituting into the Equation**:
- Substitute \( \frac{dv}{dt} \) into the equation:
\[ \frac{dv}{ds} \cdot v = \frac{v^2}{R} \]
- Dividing both sides by \( v \) (assuming \( v \neq 0 \)):
\[ \frac{dv}{ds} = \frac{v}{R} \]

6. **Separating Variables**:
- Rearranging gives:
\[ \frac{1}{v} dv = \frac{1}{R} ds \]

7. **Integrating Both Sides**:
- Integrate both sides:
\[ \int \frac{1}{v} dv = \int \frac{1}{R} ds \]
- This results in:
\[ \ln |v| = \frac{s}{R} + C \]
- Where \( C \) is the constant of integration.

8. **Applying Initial Conditions**:
- At \( s = 0 \), \( v = v_0 \):
\[ \ln |v_0| = C \]
- Thus, the equation becomes:
\[ \ln |v| = \frac{s}{R} + \ln |v_0| \]

9. **Exponentiating Both Sides**:
- Exponentiating gives:
\[ |v| = |v_0| e^{\frac{s}{R}} \]
- Since speed cannot be negative, we can drop the absolute value:
\[ v = v_0 e^{\frac{s}{R}} \]

10. **Final Expression**:
- The speed of the particle as a function of the distance covered \( s \) is:
\[ v = v_0 e^{-\frac{s}{R}} \]

### Final Answer:
\[ v = v_0 e^{-\frac{s}{R}} \]