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A conveyor belt $8.00 \mathrm{m}$ long moves at $0.25 \mathrm{m} / \mathrm{s}$. If a package is placed at one end, find its displacement from the other end as a function of time.

$s=8.00-0.25 t$

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 1

Applications of the Indefinite Integral

Harvey Mudd College

Baylor University

University of Nottingham

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So we're told that this conveyor belt here, I'll just draw a quick sketch of, it's going to be, our conveyor belt is eight m long And we're told that if we set something on this conveyor belt, it moves at two point or sorry, .25 meters per second. So every second we're moving .25 m. And we want to figure out a equation for the displacement from the other side of our conveyor belt. So that means that initially are displacement is going to be eight m since we're eight m away from the other side of the conveyor belt when we place an object on there. So we can say that our initial to play our initial displacement, The zero is equal to eight m. And so our displacement equation is just going to be our initial displacement and then plus um our velocity. So in this case it's negative .25 m per second Times T. And the reason it's negative .25 m is because we're getting closer and closer to the other side of our compare belt, so our displacement is actually decreasing as time goes on Since we start with initial displacement of eight m from the other side of our conveyor belt. And then we're moving towards our conveyor belt at .25m/s. So our displacement gets um minus, or we get closer by 0.25 m every second. So that's why our displacement is going to be equal to our initial displacement plus negative 0.25 times T. So then we can just say our displacement equation is equal to 8 -25 times t. And again, the reason that we have this minus 0.25 times T. Is because we're going towards the other side of the conveyor belt At a speed of .25 m/s.

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