A rocket is moving vertically upward against gravity. Its mass at time t is `m=m_0-mut` and it expels burnt fuel at a speed u vertically downward relative to the rocket. Derive the equation of motion of the rocket but do not solve it. Here, `mu` is constant.

Explanation

To derive the equation of motion for the rocket, we will follow these steps:

### Step 1: Define the mass of the rocket
The mass of the rocket at time \( t \) is given by:
\[ m = m_0 - \mu t \]
where \( m_0 \) is the initial mass of the rocket and \( \mu \) is the rate at which the mass of the rocket decreases due to the expulsion of burnt fuel.

**Hint:** Remember that the mass decreases linearly with time due to the fuel being expelled.

### Step 2: Identify forces acting on the rocket
The forces acting on the rocket are:
1. The thrust force \( F_t \) acting upwards due to the expulsion of fuel.
2. The weight \( W \) acting downwards due to gravity, which is given by \( W = mg \).

The net force \( F_{net} \) acting on the rocket can be expressed as:
\[ F_{net} = F_t - W \]

**Hint:** Consider how the thrust force is generated by the expelled fuel and how it relates to the rocket's motion.

### Step 3: Express the thrust force
The thrust force can be expressed in terms of the velocity of the expelled fuel \( u \) and the rate of change of mass \( \frac{dm}{dt} \):
\[ F_t = u \left( -\frac{dm}{dt} \right) \]
Here, \( -\frac{dm}{dt} \) is the rate at which mass is expelled (which is positive since mass is decreasing).

**Hint:** The thrust force is related to the velocity of the expelled fuel and the rate of mass loss.

### Step 4: Write the equation of motion using Newton's second law
According to Newton's second law:
\[ F_{net} = m \frac{dv}{dt} \]
Substituting for \( F_{net} \):
\[ F_t - mg = m \frac{dv}{dt} \]

**Hint:** Use the relationship between net force and acceleration to set up the equation.

### Step 5: Substitute the expressions for thrust and weight
Substituting the expressions for thrust and weight into the equation gives:
\[ u \left( -\frac{dm}{dt} \right) - mg = m \frac{dv}{dt} \]
Substituting \( m = m_0 - \mu t \):
\[ u \left( -\frac{dm}{dt} \right) - (m_0 - \mu t)g = (m_0 - \mu t) \frac{dv}{dt} \]

**Hint:** Ensure to replace the mass in the equation with its expression in terms of \( m_0 \) and \( \mu t \).

### Step 6: Express \( \frac{dm}{dt} \)
Since \( m = m_0 - \mu t \), we have:
\[ \frac{dm}{dt} = -\mu \]
Substituting this into the equation gives:
\[ -u(-\mu) - (m_0 - \mu t)g = (m_0 - \mu t) \frac{dv}{dt} \]

**Hint:** This substitution simplifies the equation significantly.

### Step 7: Rearrange the equation
Rearranging the equation gives:
\[ u\mu - (m_0 - \mu t)g = (m_0 - \mu t) \frac{dv}{dt} \]

**Hint:** Isolate \( \frac{dv}{dt} \) to express it in terms of the other variables.

### Final Equation
The final form of the equation of motion is:
\[ (m_0 - \mu t) \frac{dv}{dt} = u\mu - (m_0 - \mu t)g \]

This is the required equation of motion for the rocket.