Air resistance acting on a falling body can be taken into account by the approximate relation for the acceleration:
a=dv/dt=g−kv,
where k is a constant. Derive a formula for the velocity of the body as a function of time assuming it starts from rest (v=0 at t=0). [*Hint*: Change variables by setting u=g−kv.]

Express your answer in terms of the variables k, t, and appropriate constants. Determine an expression for the terminal velocity, which is the maximum value the velocity reaches.

Express your answer in terms of the variable k and appropriate constants.

To derive the formula for the velocity of the body as a function of time under the influence of air resistance, we start with the given differential equation:

\[
a = \frac{dv}{dt} = g - kv
\]

where:
- \( g \) is the acceleration due to gravity,
- \( k \) is a constant related to air resistance,
- \( v \) is the velocity of the body.

### Step 1: Introduce a substitution

As suggested, we introduce a new variable \( u \) defined as:

\[
u = g - kv
\]

This implies:

\[
\frac{du}{dt} = -k \frac{dv}{dt}
\]

Since \( \frac{dv}{dt} = g - kv = u \), we substitute:

\[
\frac{du}{dt} = -ku
\]

### Step 2: Solve the differential equation for \( u(t) \)

This is a separable differential equation. We can rewrite it as:

\[
\frac{du}{u} = -k dt
\]

Integrating both sides:

\[
\ln |u| = -kt + C
\]

Exponentiating both sides to solve for \( u \):

\[
u = C e^{-kt}
\]

where \( C \) is a constant of integration. Since \( u = g - kv \), we have:

\[
g - kv = C e^{-kt}
\]

### Step 3: Apply initial conditions

At \( t = 0 \), the initial velocity \( v(0) = 0 \), so:

\[
g - k(0) = g = C e^0
\]

This gives us \( C = g \). Therefore, the equation for \( u \) becomes:

\[
g - kv = g e^{-kt}
\]

### Step 4: Solve for \( v(t) \)

Now, solve for \( v(t) \):

\[
kv = g(1 - e^{-kt})
\]

So, the velocity as a function of time is:

\[
v(t) = \frac{g}{k}(1 - e^{-kt})
\]

### Step 5: Determine the terminal velocity

The terminal velocity is the maximum value that \( v(t) \) can reach, which occurs as \( t \) approaches infinity. As \( t \to \infty \), \( e^{-kt} \to 0 \), so:

\[
v_{\text{terminal}} = \lim_{t \to \infty} v(t) = \frac{g}{k}(1 - 0) = \frac{g}{k}
\]

### Final Answer

- The velocity as a function of time is:

\[
v(t) = \frac{g}{k}(1 - e^{-kt})
\]

- The terminal velocity is:

\[
v_{\text{terminal}} = \frac{g}{k}
\]

---

Would you like further clarification on any step, or do you have other questions? Here are 5 related questions to expand on this topic:

1. How does the velocity-time graph for this scenario look, and what does it tell you about the body's motion?
2. What would happen if the object had an initial velocity different from zero?
3. How does the terminal velocity change with different values of \( k \)?
4. How would the solution change if the air resistance was proportional to the square of the velocity instead?
5. Can you determine the time it takes for the body to reach a certain percentage of its terminal velocity?

**Tip:** In physics, recognizing when a problem can be solved using a change of variables is a powerful technique, especially when dealing with differential equations.

Learn how to derive the formula for velocity of a falling body considering air resistance, starting from rest. Includes step-by-step solution using differential equations and change of variables. Suitable for undergraduate level physics and mathematics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation