Problem

If we ignore air resistance, we can conclude that…

05:09

Problem 35 Easy Difficulty

An object with mass $ m $ is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If $ s(t) $ is the distance dropped after $ t $ seconds, then the speed is $ v = s'(t) $ and the acceleration is $ a = v'(t). $ If $ g $ is the acceleration due to gravity, them the downward force on the object is $ mg - cv, $ where $ c $ is a positive constant, and Newton's Second Law gives
$ m \frac {dv}{dt} = mg - cv $
(a) Solve this as a linear equation to show that
$ v = \frac {mg}{c} (1 - e^{-ct/m}) $
(b) What is the limiting velocity?
(c) Find the distance the object has fallen after $ t $ seconds.

Answer

a)$$v(t)=(m g / c)\left[1-e^{-(c / m) t}\right]$$
b) $$\lim _{t \rightarrow \infty} v(t)=m g / c$$
c)$$(m g / c)\left[t+(m / c) e^{-(c / m) t}\right]-m^{2} g / c^{2}$$

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 5

Linear Equations

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Video Transcript

Okay, so we're given an object off mass. Mm is dropped from a height, okay. And the distance traveled after dropping is given by S. T on the velocity of the mass we included in that process is witty and acceleration is pretty okay. And now we know by Newton's law that velocity is the rate of change off this displacement and exploration is rate of change off velocity. So we have this relation and also it is mentioned that acceleration due to gravity is g on. There is an arbitrary constants C which is limiting the floor or the movement off. The given object on that flow off movement is restricted by M g minus cv Andi In general, we call this as a drag force which act on object and limits its velocity in a resistant medium. For example, in our environment, where there is a okay, now we have to solve the situation for read. Okay, so we have to find the velocity from this Newton's law. So let's start with first one m d. V by D. T is mg minus C v. Okay, so here are dependent variable is we and independent variable iss t so right this term here so we'll have M d v by d t plus c V is equals to m g. Now we don't need any time here, so we'll divide by m so this will become d V by d. T is a close sorry plus C by M. V is close to G. Okay, now this is a form off linear differential equation. That is D let's say D y by d. X plus p y is equals toe Q. And to solve the search functions, we have an integrating factor, which is? It is to integration off p d x great. And the solution becomes what white times integrating factor is equal to integration off a few times integrating factor The X and plus arbitrary constant with decides the particular solution. Okay, so now we'll do that. So you're integrating factor will be It is toe integration off since it's p and it is multiply with dependent variable So dependent variable yer multiply Thomas CBM So CBM our i d. X So here we left DT and since the way I am is constant okay, is that arbitrary? Concert is the mass of the object and It's not wearing with time, so C by, um, will come out off integration. Sorry. Yeah, and we'll have integration off one. So that gives us it is to see by m t. Great. So we have our integrating factor now. Now what we'll do, we'll answer this directly by using this formula. So why Times integrating factor, which is here, is to see by M or T. Sorry into t is integration off Q and Q is, er, turn to the right which is not ah, function off y right. So G is a constant function, right? Its value is fixed off for a particular range if you can see so g is okay, so g times it is to see by m t d t. Now this is a simple integration we can do. It is religious constant will come out on. This is it is to a X right it is. It is to a X since ace constant value and x r is variable. So integration of this is it is to a X over a. So this gives us it is to see by m t over sea by em and we'll have a plus constant value. Okay, so now we have a solution. Sorry, I've just missed with this, So this is not why, but it's a good It's a wee. Okay, so Well, right here. We similarly here we So remember, whenever you're copping off formula, always change. Check your variables if it is correct or not. Okay, so now we have we times it is to see by empty is equals to this and will go up So m g o r c times it is to see by empty plus c. Okay, so now we can divide this value toe all these terms, So we'll have velocity since we need velocity. Right. So velocity will be mg oversee this divided by this will give us one plus. Now this becomes see over it is to see by empty. Okay, so I'll write the final solution. But we need value off. See? Right. We don't know yet. So m g or C plus, they shouldn't go up. So this will become C times. It is two minus C by M t. Okay, so now we need value off, see? And it can be calculated using a condition given to us. So what's the condition at Time T equals to zero. Since the object is dropped, velocity will be zero, so we'll substitute it. So velocity zero mg by sea will remain as it is plus C times Here is two minus C by m times t zero So this is zero is equals to m G o R c. Remember this c and this is not same. So you can give another variable name. Let's say C one Okay, I'll change it so that you don't have confusion. This c is a arbitrary constant that came due to integration. And this sea is given to her given into a question on which is ah, limiting factor to increasing velocity due to gravity or resistance that limits it. Okay, so see one again. This is and this is C one. This is C one and this is C one. Okay, so m g o S c plus C one it is +20 is one. So what will get a C one is minus M g. Oh, I see. Okay, so now our velocity will be mg. We had Yeah, M G or C plus C one, which is minus M g. Who will see it is to minus C by M t. So if you take mg will see common, we'll have one minus it is two minus C by M. T. And that's a velocity. Okay, so we are done with first question. The second is asking for limiting velocity. Okay, So limiting velocity will be when? When our time is infinity. So we'll apply formula limit when time it really tends to infinity. What is my velocity? So that becomes they meet Time t tends to infinity off mg oversee one minus. It is two minus C by M. T. Right. And we know the constant values limit doesn't change because if you have any constant and you move in a time, it will remain as it is. So m g was he one minus? And when we substitute infinity here, this becomes serious to minus infinity. Correct. And it is two minus infinity's zero. Good. So m g o s c one minus zero. So are limiting with velocities. M g o c. Okay, so we have a limiting velocity. I remember why this is coming because off the drag or what we can say a condition given that velocity is limited with some variable which is resisting this MGs velocity in agreement. Okay, great. So he had done with this. Now 30 is asking for what is distance traveled with respect toe time. So what? We have other velocity. Velocity is mg who was see one minus It is two minus C by m. T. And we know velocity is what rate or change off displacement. So D s by D t is m g o c. One minus it is two minus C by m. T. Now we can integrate it directly so we'll have the yeses equals toe m g o R c one minus. Here is two minus c by empty DT. Okay. And we can integrate both side. So what we'll have is s which is function off time. Mg by sea is constant. Okay, One integrationist e minus. Again. Integration of this will be It is two minus C by m. T O minus. Eva m bless and take C two. Since we have already done with C one. Okay, so if you simplify this, we get SS m g o s. C Okay. And this becomes de and minus you buy milk and go up and we'll go up. So we left plus minus minus. Plus M O R c. It is two minus C by M de plus C two. Okay, so now again, this is a general solution and we need value off. See, too. So we'll apply again our condition at time t close to zero. Since we're starting and we are let's say we have this, okay and object is drawn from here. Let's say this is ground and this is the top position. And from the air, object is drawn and distance measure is from this point, this is a distance. So at time T equals to zero. A displacement will be zero in this case, displacement and distances. Same because object is traveling in a straight line. Okay, and it's not changing it, but okay, so we'll have zero is equals two mg by sea, t zero plus m by sea. It is too minus. If I'm in 20 is zero plus C two. So this gives us zero is mg by sea times it is. Zero is one. So this is NBC plus C two. So what will be C two c two will be m squared G o a C square on in minus. And since it will go here. Okay, so now we have our C two. So r s will become, Therefore are S S M G or C D plus. Mm bye. See? It is two minus e by empty minus M squared G over C square, and that's a solution for yes.

Pawan Y.

Mumbai university

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 5

Linear Equations

Problem

If we ignore air resistance, we can conclude that…

Problem 35 Easy Difficulty

Answer

a)$$v(t)=(m g / c)\left[1-e^{-(c / m) t}\right]$$
b) $$\lim _{t \rightarrow \infty} v(t)=m g / c$$
c)$$(m g / c)\left[t+(m / c) e^{-(c / m) t}\right]-m^{2} g / c^{2}$$

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Caleb E.

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Integration Review - Intro

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Watch More Solved Questions in Chapter 9

Video Transcript

Calculus: Early Transcendentals

Caleb E.

Kristen K.

Michael J.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

a)$$v(t)=(m g / c)\left[1-e^{-(c / m) t}\right]$$b) $$\lim _{t \rightarrow \infty} v(t)=m g / c$$c)$$(m g / c)\left[t+(m / c) e^{-(c / m) t}\right]-m^{2} g / c^{2}$$

Calculus: Early Transcendentals

Caleb E.

Kristen K.

Michael J.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Caleb E.

Kristen K.

Michael J.

Joseph L.

a)$$v(t)=(m g / c)\left[1-e^{-(c / m) t}\right]$$
b) $$\lim _{t \rightarrow \infty} v(t)=m g / c$$
c)$$(m g / c)\left[t+(m / c) e^{-(c / m) t}\right]-m^{2} g / c^{2}$$