Airline travel The following figure shows the position funct...

Question

Airline travel The following figure shows the position function of an airliner on an out-and-back trip from Seattle to Minneapolis, where s = f(t) is the number of ground miles from Seattle t hours after take-off at 6:00 A.M. The plane returns to Seattle 8.5 hours later at 2:30 P.M. <IMAGE>
d. Determine the velocity of the airliner at noon (t = 6) and explain why the velocity is negative.

d. Determine the velocity of the airliner at noon (t = 6) and explain why the velocity is negative.

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A car starts from city A and travels towards city B. The position function of the car is given by S equals GFT, where S is the distance in kilometers from city A. And tea is the time and hour since departure at 8:00 a.m. The car makes a return trip to C A, arriving back 8.5 hours later at 4:30 p.m. Approximate the velocity of the car at 2 p.m. and state the reason for the velocity being negative. Awesome. So it appears for this particular prompt we're asked to solve for two separate answers. We're trying to figure out the velocity because we're asked once again, we're trying to figure out the velocity of the car at 2 p.m. and we're also asked to state the reason for why the velocity is a negative value. So those are our two answers that we're ultimately trying to solve for. So before we start solving for this problem, now that we know what we're trying to solve for, let's take a moment to investigate our figure that is provided to us. As you can see, the Y axis denotes S, our position, which is going to be in units of kilometers from city A, and then our X axis denotes our time in units of hours T. Fantastic. And as you can see, our curve for our position function starts at 00, the origin point, and increases until it peaks at its highest peak. Then it looks like it remains constant, neither increasing nor decreasing for a period of time until it starts to decrease and trend back towards the T axis, which represents our hours. Awesome. So now that we have a visualization of what's happening for this particular prom, let us recall a note once again that at 2 p.m. Which means that at 2 p.m., so if you look at 20 p.m., that means that tea will equal 6 hours, so, T equals 6 hours. So that will mean that the car is on its return trip to City A. So at 2 p.m. we need to note that T equals 6 hours and the car will be on its return trip back to City A. And we also need to recall and note that the graph GF T is approximately linear, so the graph of GT will be approximately linear at parentheses 6.7. And this is the interval that is close to T equals 6, so therefore it will be a good approximation to use for the velocity at T equals 6. So, therefore, in order to estimate of 6, we need to recall and use the average velocity formula between T equals 6 and T equals 7. So when we do that, as we should recall, the average will be equal to G 7 minus G of 6, all divided by 7 minus 6, so that will mean that the average. It's going to be equal to, so let's determine what G of 7 is. So that's essentially saying that X is equal to 7. So in this case, the X axis denoted by T, so we go to T equals 7 and we go up and we'll find that G of 7. It's going to be equal to 75 kilometers, and G of 6, so going to where T equals 6, and we go up, we will find that GF 6 is equal to 150 kilometers. So when we plug in our known variables, we will take 75 kilometers minus 150 kilometers, all divided by. One is 7 minus 6 is 1, so that will mean that the average is equal to when we plug this into a calculator, negative 75 kilometers per hour. And hooray, we did it. We found our V average. So now, since we know that GFT is decreasing after T equals 5, that will mean that the velocity is negative, which will indicate that there is motion going towards city A. So that means without a doubt, That our final answers have to be Once again, to quickly recap, our final answers have to be the velocity at T equals 62 p.m. will be approximately 75 kilometers per hour, and the negative sign means that the car is moving back towards city A. And that's it. We've officially solved for this problem. Hooray. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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