A particle initially at rest is subjected to two forces, one is constant, the other is retarding force proportional to the particle velocity. In the subsequent motion of the particle, the acceleration

Options

A. will increases then decreases
B. will decreases than increases
C. will decreases to zero
D. will decrease and become constant at some non zero value.

Answer

To solve the problem, we need to analyze the motion of a particle subjected to two forces: a constant force \( P \) and a retarding force \( F_r \) that is proportional to the velocity \( v \) of the particle. The retarding force can be expressed as \( F_r = k \cdot v \), where \( k \) is a constant.

### Step-by-Step Solution

1. **Identify the Forces**:
- The particle is subjected to a constant force \( P \) acting in the positive direction.
- The retarding force \( F_r \) is given by \( F_r = k \cdot v \), acting in the opposite direction of the motion.

2. **Write the Equation of Motion**:
- According to Newton's second law, the net force acting on the particle is equal to the mass \( m \) of the particle multiplied by its acceleration \( a \):
\[ m \cdot a = P - F_r \]
Substituting \( F_r \):
\[ m \cdot a = P - k \cdot v \]

3. **Express Acceleration in Terms of Velocity**:
- We know that acceleration \( a \) can also be expressed as \( \frac{dv}{dt} \):
\[ m \cdot \frac{dv}{dt} = P - k \cdot v \]

4. **Rearrange the Equation**:
- Rearranging gives:
\[ \frac{dv}{P - k \cdot v} = \frac{dt}{m} \]

5. **Integrate Both Sides**:
- Integrate the left side from \( v = 0 \) to \( v \) and the right side from \( t = 0 \) to \( t \):
\[ \int_0^v \frac{1}{P - k \cdot v} \, dv = \int_0^t \frac{1}{m} \, dt \]
- The left side integrates to:
\[ -\frac{1}{k} \ln |P - k \cdot v| \bigg|_0^v = \frac{t}{m} \]

6. **Solve for Velocity**:
- After integrating and simplifying, we find:
\[ -\frac{1}{k} \left( \ln |P - k \cdot v| - \ln |P| \right) = \frac{t}{m} \]
- This can be rearranged to:
\[ P - k \cdot v = P \cdot e^{-\frac{kt}{m}} \]
- Thus, solving for \( v \):
\[ k \cdot v = P - P \cdot e^{-\frac{kt}{m}} \]
\[ v = \frac{P}{k} (1 - e^{-\frac{kt}{m}}) \]

7. **Find Acceleration**:
- To find acceleration, differentiate \( v \) with respect to \( t \):
\[ a = \frac{dv}{dt} = \frac{P}{k} \cdot \frac{k}{m} e^{-\frac{kt}{m}} = \frac{P}{m} e^{-\frac{kt}{m}} \]

8. **Behavior as Time Increases**:
- As \( t \) approaches infinity, \( e^{-\frac{kt}{m}} \) approaches 0, thus:
\[ a \to 0 \]
- This indicates that the acceleration decreases over time and approaches zero.

### Conclusion
The acceleration of the particle decreases exponentially and approaches zero as time tends to infinity.