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A falling object of mass $m$ can reach terminal velocity when the drag force is proportional to its velocity, with proportionality constant $k$ . Set up the differential equation and solve for the velocity given an initial velocity of $0 .$

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

Calculus 2 / BC

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

Differential Equations

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Lectures

01:11

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Linear drag Derive the equ…

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Constant acceleration Supp…

02:03

05:12

According to one model tha…

02:22

Moving through a liquid, a…

01:00

What is the acceleration o…

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A body of mass $m$ is proj…

02:43

Velocity in a Resisting Me…

04:38

(III) Air resistance actin…

question wants us to prove that the acceleration of this falling body is constant. When the velocity is K time squared of S meters per second or Kay's a constant and s is the meters it's fallen from its starting point. So we want to be solving for TV DT because that's acceleration and were given s here, not tea. So we need to use the chain rule and find D V D s and then use d S D team. So by doing this, we can rewrite this as t s g t Right There you go. It's pretty ugly. We want to rewrite this as Devi. Yes, this is the same. But the STT can be re written as velocity because that's what it is. It's the change and ah, our distance over a change of time. So we're just multiplying velocity by the change in the change of last year the change in s And so if we do this, we do this out, we can just plug in the first we know it's k square root of ass, right? And then we can do DVDs and if we take the derivative of the with respect to s we get? Okay over. Well, she OK over two squares. Of s because we have this 1/2. That's the 1/2 power we concert like that. Bring it down and we have to the negative 1/2 power. So they get k times to overrule us. And that gives us k squared over two. And that is our acceleration and its constant. Because kay is a constant. So no matter what happens, the velocity acceleration is always going to be a case where it over too.

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