We are tasked with solving the first-order linear differential equation:

$m \frac{dv}{dt} = mg - kv$

where:

• $m$ is the mass of the object,
• $g$ is the acceleration due to gravity,
• $k$ is the proportionality constant of the air resistance (friction),
• $v(t)$ is the velocity as a function of time.

Additionally, we have the initial condition: $v(0) = 0$.

Step 1: Rearranging the Equation

First, we can rewrite the differential equation as:

$\frac{dv}{dt} = g - \frac{k}{m} v$

Step 2: Solving the Differential Equation

This is a first-order linear differential equation. To solve it, we can use the method of integrating factors.

The equation can be written in the form:

$\frac{dv}{dt} + \frac{k}{m} v = g$

Now, the integrating factor $\mu(t)$ is given by:

$\mu(t) = e^{\int \frac{k}{m} dt} = e^{\frac{k}{m}t}$

Multiplying both sides of the differential equation by $\mu(t)$:

$e^{\frac{k}{m}t} \frac{dv}{dt} + e^{\frac{k}{m}t} \frac{k}{m} v = g e^{\frac{k}{m}t}$

This simplifies to:

$\frac{d}{dt} \left( e^{\frac{k}{m}t} v \right) = g e^{\frac{k}{m}t}$

Step 3: Integrating Both Sides

Now, integrate both sides with respect to $t$:

$e^{\frac{k}{m}t} v = \int g e^{\frac{k}{m}t} dt$

The right-hand side is:

$\int g e^{\frac{k}{m}t} dt = \frac{g m}{k} e^{\frac{k}{m}t}$

So, we have:

$e^{\frac{k}{m}t} v = \frac{g m}{k} e^{\frac{k}{m}t} + C$

where $C$ is the constant of integration. Dividing through by $e^{\frac{k}{m}t}$, we get:

$v = \frac{g m}{k} + C e^{-\frac{k}{m}t}$

Step 4: Applying the Initial Condition

To determine $C$, we use the initial condition $v(0) = 0$:

$0 = \frac{g m}{k} + C e^{0}$

This gives:

$0 = \frac{g m}{k} + C \quad \Rightarrow \quad C = -\frac{g m}{k}$

Step 5: Final Formula for $v(t)$

Substitute the value of $C$ into the general solution:

$v(t) = \frac{g m}{k} - \frac{g m}{k} e^{-\frac{k}{m}t}$

Factor out $\frac{g m}{k}$:

$v(t) = \frac{g m}{k} \left( 1 - e^{-\frac{k}{m}t} \right)$

Thus, the formula for the velocity $v(t)$ during the period of the object's fall is:

$v(t) = \frac{g m}{k} \left( 1 - e^{-\frac{k}{m}t} \right)$

Interpretation:

• As $t \to \infty$, $v(t) \to \frac{g m}{k}$, which is the terminal velocity.
• For small $t$, the speed starts at 0 and gradually increases.

Would you like details or have any questions on this? Here are 5 related questions:

1. How can you determine the terminal velocity of an object in free fall with air resistance?
2. What happens to the velocity if the friction constant $k$ increases?
3. How would the solution change if the initial velocity wasn't zero?
4. Can you derive the time it takes to reach half of the terminal velocity?
5. How does the mass $m$ of the object affect its velocity over time?

Tip: In many problems involving air resistance, terminal velocity is a key concept that helps determine when forces reach equilibrium.