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Solve the given problems by integration.A marathon runner's speed (in $\mathrm{km} / \mathrm{h}$ ) is $v=\frac{12.0}{0.200 t+1} \cdot$ How far does the runner go in $3.00 \mathrm{h}$ ?

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 2

The Basic Logarithmic Form

Integrals

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were given functions for the velocity of two runners. The first is aged, represented by a of tea, and the second is Bob, represented by B of tea. And we want to know who is ahead when t is five and who's ahead 20 is 10 and then we want to graft these two position functions that we find and see which one is bounded and thus which velocity will be finite. So we'll start with part a the position. If we have the velocity, the position is given by the integral oh velocity. So when we want Abe's position at five of tea 30 we'll take the integral from zero to five. So the initial to what we're asked about and they will put in our Abe function four over t plus one t t the Ford in come out. We can recognize that this is one over function and so we'll get the natural long so it will be four Ln t plus one. When we take the integral, this is evaluated from 0 to 5, the natural log of 10 So we're only gonna maintain our four Ln of five plus one is six. We put that into a calculator will get 7.17 miles is the position. Now we'll do the same for B. The integral from 0 to 5 of our bob function is for e to the negative t over to d. T no, this fork and come out as a constant and the negative t over to. We'll get negative two multiplied by our outcome and so will look like this Negative eight e to the negative t over to that's evaluated from 0 to 5. We don't want that negative in there, so we'll swap the bounds. That will be positive. Eight e to the negative t over to from 5 to 0 to get rid of that negative value. There we do. This out will get eaten. Zero is one. And so this will be eight minus. When we put in five for tea, we get eight e to the negative 5/2. We could use a calculator for this and get 7.34 miles as the position, and that's greater than 7.17 So Bob is ahead after tea is five. Now we need to do the same for 10. So a 10 is equal to the integral from 0 to 10 of our Abe function for over T plus one d t. This is the same thing with different bounds. So four natural log of T plus one from 0 to 10 again, the natural log of one is going to be zero. So we just have four Ln 11 for 10 plus one. We use a calculator and get 9.59 miles. Be of 10 to the position of Bob at 10 will be the integral from 0 to 10 of four e to the negative. T over to D. T. This is again will be the same as before. So eight eat negative t over to from 10 to 0. We're getting rid of that. Negative. I will get that. This is eight minus eight e to the negative five. Turnover two is five. That will be 7.95 miles. As might have guessed, this is swamped. And so now aid is ahead. That's part of it. Now we're gonna move on to Part B. We want to make sure that we maintain our initial well, our initial point of 00 the origin so first, we're just taking the general integral of what we did before. So we're not using any bounds. So this will be the integral for aid would be the integral of four over t plus one D t. And this will be for tens natural log of t plus one plus c constant. Because it's an indefinite integral. We need to remember to do that. We'll do the same for B. The integral of four e to the negative. T over to D. T is equal to negative eight e two negative t over two plus C now is when we use our initial value. So we want if his tea is zero, we want our A function here to be zero. So if t zero we have a natural log of one, which is zero So we're good there. We get that c will go to zero because we don't need to add a constant. And then we also want if t is here here for our function to be zero. So if we have t zero, we get e to the zero, which will be once we have negative eight. So see will be positive eight contract that to get zero. And now we're gonna graft these two. So we'll make a little graph there and using a graphing calculator and simplifying our be function or bob function to be eight times one minus e to the negative t over to. We can just do this by transformations. But we're just gonna use a graphing calculator here, and our aid function looks like that. And our bob function looks like this. And this, Ahsan told here that we're noticing is at eight. So we'll notice that a here is unbounded and be here for Bob is bounded by eight. This is clearly a little closer to reality because people can't run infinitely fast.

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