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(III) Air resistance acting on a falling body can be taken into account by the approximate relation for the acceleration: $a=\frac{d v}{d t}=g-k v$ where $k$ is a constant. (a) Derive a formula for the velocityof the body as a function of time assuming it starts from rest $(v=0$ at $t=0) .$ [Hint. Change variables by setting $u=g-k v . ](b)$ Determine an expression for the terminal velocity, which is the maximum value the velocity reaches.

a) $\frac{g\left(1-e^{-k t}\right)}{k}$b) $\frac{g}{k}$

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Cornell University

Rutgers, The State University of New Jersey

University of Winnipeg

Lectures

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

04:16

In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

03:53

(III) Air resistance actin…

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22. Air resistance acting …

10:07

Problem 4 (b) Let Uo be th…

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An object with mass $m$ is…

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An object with mass $ m $ …

05:03

01:02

08:29

. Let $v(t)$ be the veloci…

07:48

An object falling from res…

07:05

A differential equation g…

so here for part A. We know that the velocity is found by integrating the acceleration with respect the time and so we can use the substitution. And so we can say that the initial value of you so you initial is gonna be G minus K V initial. And this is an equal country. So we can say that a The acceleration is going to be equal to the derivative of the velocity with respect to time. Therefore, the V will equal a d t ah, and you can say d v equals G minus k v times d t. And so we consider a d the over G minus k v will equal d t and then we can make the substitution where you equals g minus k v. And so we have that Devi equals negative D'You divided by K. Let's get a new workbook here we have Devi divided by G minus K V. This is equaling again d t. And we can say negative d'you over k times one over you, uh, is going to equal d t. And we can say, of course, do you over you is gonna equal negative k d t and now we finally integrate. So integrate this from G to you and then integrate this from of course, zero to t. This is an equal negative K Times zero to t of d t. And we finally have natural log of you evaluated at G and at you is gonna equal negative k t. Uh, let's get a new workbook. So u s so we can say the natural law of you over G. It's an equal negative, Katie. So we can say that you is gonna be equal to G times. Uh, e to the negative. Katie, remember, E is simply just the ah value 2.7 three approximately so g to the e uh, g to the e to the negative Katie Power. And, uh, this is gonna equal g minus K. V. Um, basically, we're substituting back in for you. This right here. So this is when we substitute back in for you and then solve for v So isolated Viet This point So v unequal G over K times one minus e to the negative k t power. So this would be your answer for party for part B. Now we want to find the terminal velocity so the velocity terminal would be equal to the limit. Well, to the limit as T approaches infinity of G over K to the one minus E to the negative. Katie Power. And this is simply going to This is an exponential decay, of course. So the terminal velocity would simply be equal to G. Okay. Ah, we also just to note that if acceleration was zero, uh, which again happens that terminal velocity than A is gonna equal g minus k V and this is gonna equal zero. And then we can simply solved for V. So these gonna equal g over K. So if we say that the acceleration is zero at terminal velocity, then we can just set up an equation. Sulfur v. However, this way is a lot better because we can. This gives you more and for me a good see more information in the sense that velocity the definition for the for the terminal velocity is the limit. As time approaches infinity eso again the terminal velocity will equal g over. Okay, this is your answer for part B. That is the end of the solution. Thank you for watching

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