College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Understanding Motion with Air Resistance**
When you drop an object of mass \( m \) from a tall building, the primary forces affecting its motion are gravity and air resistance. Air resistance is proportional to the object's speed with a positive constant of proportionality \( k \). Let \( g \) represent gravitational acceleration (a positive constant).
**Expressing Total Force**
To express the total force \( F \) in terms of \( m \), \( g \), and the object's velocity \( v \) (considering upward displacement as positive), the equation is:
\[
F = mg - kv
\]
**Newton’s Second Law**
According to Newton’s second law, force equals mass times acceleration, expressed as \( F = ma \). By relating acceleration to velocity, we can rewrite the total force equation as a first-order differential equation for \( v \) as a function of time \( t \). Denote acceleration \( v' \) as \( \frac{dv}{dt} \).
**Differential Equation**
To solve the differential equation,
\[
m \left(-\frac{dv}{dt}\right)
\]
results in a syntax error indicating that this is not an equation as presented.
**Solving for Velocity**
Given the initial condition \( v(0) = v_0 \), solving the differential equation gives:
\[
v(t) = \frac{mg}{k} \left(1 - e^{-\frac{k}{m}t}\right)
\]
**Finding Terminal Velocity**
The terminal velocity is determined when the object's speed stops changing:
\[
\text{Terminal velocity} = \frac{mg}{k}
\]
*Note: The boxes above indicate where certain expressions need correction.*](https://content.bartleby.com/qna-images/question/f57c5c9b-acb3-4a3d-a39b-55eeff7ee4ad/abf29cbb-a8ae-439f-86ca-2d69bc534699/v4n4as_processed.jpeg)
Transcribed Image Text:**Understanding Motion with Air Resistance**
When you drop an object of mass \( m \) from a tall building, the primary forces affecting its motion are gravity and air resistance. Air resistance is proportional to the object's speed with a positive constant of proportionality \( k \). Let \( g \) represent gravitational acceleration (a positive constant).
**Expressing Total Force**
To express the total force \( F \) in terms of \( m \), \( g \), and the object's velocity \( v \) (considering upward displacement as positive), the equation is:
\[
F = mg - kv
\]
**Newton’s Second Law**
According to Newton’s second law, force equals mass times acceleration, expressed as \( F = ma \). By relating acceleration to velocity, we can rewrite the total force equation as a first-order differential equation for \( v \) as a function of time \( t \). Denote acceleration \( v' \) as \( \frac{dv}{dt} \).
**Differential Equation**
To solve the differential equation,
\[
m \left(-\frac{dv}{dt}\right)
\]
results in a syntax error indicating that this is not an equation as presented.
**Solving for Velocity**
Given the initial condition \( v(0) = v_0 \), solving the differential equation gives:
\[
v(t) = \frac{mg}{k} \left(1 - e^{-\frac{k}{m}t}\right)
\]
**Finding Terminal Velocity**
The terminal velocity is determined when the object's speed stops changing:
\[
\text{Terminal velocity} = \frac{mg}{k}
\]
*Note: The boxes above indicate where certain expressions need correction.*
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