Solution Found!
1.3 we saw that the autonomous differential equation where
Chapter 2, Problem 40E(choose chapter or problem)
Terminal Velocity In Section 1.3 we saw that the autonomous differential equation
\(m \frac{d v}{d t}=m g-k v\)
where k is a positive constant and g is the acceleration due to gravity, is a model for the velocity v of a body of mass m that is falling under the influence of gravity. Because the term -kv represents air resistance, the velocity of a body falling from a great height does not increase without bound as time t increases. Use a phase portrait of the differential equation to find the limiting, or terminal, velocity of the body. Explain your reasoning.
Text Transcription:
m dv/dt=mg-kv
Questions & Answers
QUESTION:
Terminal Velocity In Section 1.3 we saw that the autonomous differential equation
\(m \frac{d v}{d t}=m g-k v\)
where k is a positive constant and g is the acceleration due to gravity, is a model for the velocity v of a body of mass m that is falling under the influence of gravity. Because the term -kv represents air resistance, the velocity of a body falling from a great height does not increase without bound as time t increases. Use a phase portrait of the differential equation to find the limiting, or terminal, velocity of the body. Explain your reasoning.
Text Transcription:
m dv/dt=mg-kv
ANSWER:Step 1 of 5
Given that
We have to use a phase portrait of the differential equation to find the limiting, or terminal, velocity of the body and we have to explain the reason for this.