If M is mass of rocket, r is rate of ejection of gases and u is velocity of gases with respect to rocket then acceleration of the rocket dv/dt is equal to

Options

A. `(ru)/(M -rt)`
B. `(M-rt)/(ru)`
C. `(ru)/(M+rt)`
D. `(ru)/(M)`

Explanation

To find the acceleration of the rocket, we can use the concept of thrust force and the relationship between mass, velocity, and acceleration. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \( M \) be the initial mass of the rocket.
- Let \( r \) be the rate of ejection of gases (mass flow rate).
- Let \( u \) be the velocity of the gases with respect to the rocket.

2. **Thrust Force**:
- The thrust force (\( F \)) generated by the rocket can be expressed as:
\[ F = u \cdot \frac{dm}{dt} \]
- Here, \( \frac{dm}{dt} \) is the rate of change of mass, which is equal to \(-r\) (since the mass is decreasing).

3. **Applying Newton's Second Law**:
- According to Newton's second law, the acceleration (\( a \)) of the rocket can be expressed as:
\[ a = \frac{F}{m} \]
- Substituting the expression for thrust force into this equation gives:
\[ a = \frac{u \cdot (-r)}{M - rt} \]
- Here, \( M - rt \) represents the mass of the rocket at time \( t \) after \( rt \) mass has been expelled.

4. **Final Expression for Acceleration**:
- Therefore, the acceleration of the rocket can be expressed as:
\[ \frac{dv}{dt} = \frac{-ur}{M - rt} \]

### Final Answer:
The acceleration of the rocket is given by:
\[ \frac{dv}{dt} = \frac{-ur}{M - rt} \]