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Solve the given problems by integration.Conditions are often such that a force proportional to the velocity tends to retard the motion of an object moving through a resisting medium. Under such conditions, the acceleration of a certain object moving down an inclined plane is given by $20-v$. This leads to the equation $t=\int \frac{d v}{20-v}$. If the object starts from rest, find the expression for the velocity as a function of time.

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 2

The Basic Logarithmic Form

Integrals

Missouri State University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

02:51

Position from velocity Con…

00:54

Given the velocity functio…

06:24

An object starts moving in…

03:30

07:35

Additional ProblemsThe…

01:02

Find the integral by using…

01:16

Express the solutions of t…

02:39

Motion with constant accel…

02:25

04:00

we are trying to show that the position function comes out the same regardless of the method that it is obtained. First, let's look at the anti derivative method, which says that the position function is nothing more than the integral of the velocity. So let's integrate our velocity function nine minus T square. The integral of that is nine T minus T cubed over three, plus a constant. Now the position function, it zeroes negative too. So if I fill in zero for the time, I should get a position of negative too. So let's see, what see is position functions negative to in time is zero by filling zero for the time that zero plus zero plus c therefore see is negative two. So the position function using the anti derivative method is 90 minus t cubed over three minus two. Okay, now, the second method is using the fundamental theorem of calculus that would tell us that the position function s A. T is equal to the starting position s and zero plus the integral from zero t of the velocity of X dx already established. The position function at zero is negative two and two. That the integral from zero to t of V of X dx. Okay, so the velocity it X would be nine minus x squared integrated. That would be nine x minus X cubed over three. And we are going to evaluate that from zero to t. So that would give us a position function of negative, too. By filling t. I have nine t minus t cubed over three. Subtract what I get when I fill in zero. But when I fell in zero, I get zero so I can see that that is the exact same thing as I had before.

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