Let v(t) be the velocity of an object of mass m in free fall...

Let $v(t)$ be the velocity of an object of mass $m$ in free fall near the earth's surface. If we assume that air resistance is proportional to $v^{2}$ then $v$ satisfies the differential equation $m \frac{d v}{d t}=-g+k v^{2}$ for some constant $k>0$. (a) $\operatorname{Set} \alpha=(g / k)^{1 / 2}$ and rewrite the differential equation as $$\frac{d v}{d t}=-\frac{k}{m}\left(\alpha^{2}-v^{2}\right)$$ Then solve using separation of variables with initial condition $v(0)=0$. (b) Show that the terminal velocity $$\lim _{t \rightarrow \infty} v(t)$$ is equal to $-\alpha$

Let $v(t)$ be the velocity of an object of mass $m$ in free fall near the earth's surface. If we assume that air resistance is proportional to $v^{2}$ then $v$ satisfies the differential equation $m \frac{d v}{d t}=-g+k v^{2}$ for some constant $k>0$.
(a) $\operatorname{Set} \alpha=(g / k)^{1 / 2}$ and rewrite the differential equation as
$$\frac{d v}{d t}=-\frac{k}{m}\left(\alpha^{2}-v^{2}\right)$$
Then solve using separation of variables with initial condition $v(0)=0$.
(b) Show that the terminal velocity $$\lim _{t \rightarrow \infty} v(t)$$ is equal to $-\alpha$

Key Concepts

Separation of Variables

Separation of variables is a technique for solving differential equations in which the variables can be algebraically rearranged so that all terms involving one variable occur on one side of the equation and all terms involving the other variable occur on the opposite side. This method simplifies the problem by reducing it to integrating two separate functions, one for each variable, making it a powerful tool in solving first-order ODEs.

Terminal Velocity

Terminal velocity refers to the steady-state speed achieved by an object when the net force acting on it becomes zero, meaning the accelerating force (such as gravity) is exactly balanced by the resistive force (such as drag). In the context of differential equations, finding the limit as time approaches infinity (or solving the equilibrium condition) reveals the terminal velocity, which represents the maximum speed the object will reach during its descent.

Newton's Second Law and Differential Equations

Newton's second law forms the basis for deriving differential equations that model motion. In many physical problems, forces acting on a body are translated into equations that express the rate of change of momentum or velocity. This approach allows us to build mathematical models of dynamic systems where acceleration is determined by the net forces, leading to ordinary differential equations (ODEs) that describe the evolution over time.

Quadratic Air Resistance

Air resistance often depends on the speed of an object, and when the resistance is proportional to the square of the velocity, it is called quadratic air resistance. This model is particularly relevant at higher speeds and is used to capture the nonlinear drag forces experienced by objects moving through a fluid, leading to differential equations that include quadratic terms in velocity.

Transcript

00:01 Okay, we have an object of mass m with velocity v, and it satisfies this differential equation for some k greater than zero.

00:09 So first we're going to substitute some variables and write this in a different way to solve it a different way.

00:19 And then we're going to solve it, and then we're going to take the limit of v.

00:23 Okay, so we have m, dv, d t, equals minus g plus kv, v squared.

00:34 And we're going to make the substitution alpha equals g over k to the one -half.

00:44 So alpha -squared is g over k, so g is k alpha -squared.

00:54 So we have mdv -d -t equals minus k alpha -squared plus k v squared.

01:04 So i'm going to factor a k out of there, and i get v -squared minus alpha -squared.

01:11 Oops, i'm going to factor out of minus k because i want it to look like theirs.

01:21 Whoops.

01:25 So i factor out of minus k, so i get alpha squared minus v squared, and now divide the m over there.

01:31 So dv, d t goes minus k over m, alpha squared minus v squared.

01:38 All right.

01:41 Now i'm going to solve it.

01:42 I'm going to separate the variables.

01:44 I get dv over alpha squared minus v squared equals minus k over m d t.

01:52 So this factors into alpha minus v and alpha plus v.

02:00 So i'm going to have to do partial fractions here.

02:03 A, alpha minus v plus b, alpha plus v, equals 1 over alpha squared minus v squared.

02:15 So a times alpha plus v plus b times alpha minus v equals one.

02:26 So if v equals alpha, then we get 2 alpha equals, oops.

02:42 If v equals alpha, we get 2 alpha a equals 1.

02:52 So a is 1 over 2 alpha...

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