An object falling from rest in air is subject not only to...

An object falling from rest in air is subject not only to gravitational force but also to air resistance. Assume that the air resistance is proportional to the velocity and acts in a direction opposite to the motion. Then the velocity of the object at time $t$ satisfies an equation of the form $$v^{\prime}=32-k v,$$ where $k$ is a positive constant and $v(0)=0 .$ Here we are measuring distance in feet and the positive direction is down. (a) Find $v(t)$ (b) Show that $v(t)$ cannot exceed $32 / k$ and that $v(t) \rightarrow$ $32 / k$ as $t \rightarrow \infty$ (c) Sketch the graph of $v$

An object falling from rest in air is subject not only to gravitational force but also to air resistance. Assume that the air resistance is proportional to the velocity and acts in a direction opposite to the motion. Then the velocity of the object at time $t$ satisfies an equation of the form $$v^{\prime}=32-k v,$$ where $k$ is a positive constant and $v(0)=0 .$ Here we are measuring distance in feet and the positive direction is down.
(a) Find $v(t)$
(b) Show that $v(t)$ cannot exceed $32 / k$ and that $v(t) \rightarrow$ $32 / k$ as $t \rightarrow \infty$
(c) Sketch the graph of $v$

Key Concepts

First-Order Linear Differential Equations

These are differential equations of the form y' + p(t)y = q(t) where the solution methodology typically involves using an integrating factor. They are central in modeling time-dependent processes, such as the motion of objects subject to forces like gravity and resistance.

Integrating Factor Method

This method is used to solve first-order linear differential equations. By multiplying the entire differential equation by an appropriately chosen integrating factor, the equation can be rewritten as the derivative of a product, which then simplifies the process of finding the solution.

Terminal Velocity

Terminal velocity refers to the condition where an object in free fall eventually reaches a constant velocity due to the balance between gravitational force and resistive forces (like air resistance). This concept is exemplified by the velocity approaching a limiting value over time in models involving proportional resistance.

Exponential Decay

Exponential decay describes a process in which a quantity decreases at a rate proportional to its current value. In the context of falling objects with air resistance, the difference between the current velocity and the terminal velocity decreases exponentially over time, leading to the velocity asymptotically approaching a constant value.

Transcript

00:01 Okay, so in this question we were given that v -dash is equal to 2 minus kv and v0 is equal 0 and that part a asked us what is the solution which is v t.

00:22 Okay, so v t can be found as simply either power minus h and t where h t, where h t, is the integral of pt and pt pt is pt is k okay okay um i'll like that later so we have eht times 32 d t okay now if i write this out we have h of t is the h of t is the integral of k t is the integral of k t which is just k t so vt can be written as e power minus k t times e kt times 32 dd um yeah i was serving now we just working you give my steps in space okay yeah if you integrate this you get e power k much t times um 32 the integral of 32 e k t is simply 32 upon k e power k t because the integral of k t is this e k t is this e kd and you divide it by k which is the differential of kd so this this times that okay e d d bough minus k t times e bough minus k t is just one so you have 32 k plus c is equal v t now we know v 0 is equal 0 so this can now be rewritten at oh wait e minus k t sorry yeah now this can be rewritten as 0 is equal to 32 k plus e power 0 is just 1 so this ends up being c and c is now 32 upon k if we rewrite this with c in the formula now, we will get 32k minus 32k upon k, e power minus kt.

03:21 You can take b 32 upon k common and you get 1 minus e power minus kt.

03:31 That's part a.

03:33 Part a was asking us what the solution to this is and we just found the solution.

03:37 Now part d is a little bit more tricky.

03:43 It asks us, what is part d? it asks us that we have to show that vt cannot exert 32 upon k and that as they approaches, 32 upon k, t is equal as infinity.

04:04 Okay, this is pretty easy too.

04:06 So vt, we know is 32 upon k.

04:10 1 minus e minus kt okay now if you were to for example approach 32k let's see what happens if vt is 32k right we get one minus e power minus k t if we approach is that you get one is equal one minus e power minus k t and this would just be equal to e power minus k zero which is just e power minus k k t is equal to zero now the thing is nothing nothing to the power something can give you zero that's a fundamental law so if you were to rewrite this right you get e power k t is equal to zero now this approach is zero it doesn't actually equal to zero since nothing to the power of anything can actually give you zero so since e power minus k t approaches zero it can be n -sad that since t is in the numerator of this denominator, i mean, of this fraction, then t approaches infinity, because the larger number gets in the numerator, the smaller the number will get, and the smallest positive number we can get is zero.

05:33 So it is ends proven that as e power minus kt approaches zero, t approaches infinity...

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