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Problem 17

The profits of a small company for each of the fi…

Problem 16

The accompanying figure shows the plot of distance fallen versus time for an object that fell from the lunar landing module a distance 80 $\mathrm{m}$ to the surface of the moon.
a. Estimate the slopes of the secants $P Q_{1}, P Q_{2}, P Q_{3},$ and $P Q_{4}$ arranging them in a table like the one in Figure 2.6 .
b. About how fast was the object going when it hit the surface?


For part a) see explanation for result. For part b) 16 $\mathrm{m} / \mathrm{s}$


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Video Transcript

Hello, everyone. Today we're dealing with a problem of finding slopes for secret lines. We're giving a plot of the trajectory of the object that falls from some landing module towards the surface of the moon with the wind. So to find the slope of a secret line first we start with two points and we're trying to find this slope right here. The slope is just how much we go up by bad by how much we go, right? Bye. This is the rise in this throne. So we rise up and run. Right? So this slope is just equal to the rise. Divide by the run, simple as that. So we'll start with a and the slope of peak you won is just how much we go about it. Which is a T minus 2060. So 80 minus 20. And how much we go right by. This is just 10 minus five. So I hear 10 minutes five. This gives us Sorry. 20. This gives us 60 over five, and I'm gonna put in, um, the units here, which will make sense later on. This gives us 12 meters per second for peak. You too. We similarly have 80 80 minus 38 divided by 10 months. Seven which gives us 42 over three meters per second, which is, uh, 14 meters per second. For pick you three will have 80 minus, uh 57 divide by 10 minus 8.5, which gives us 23 divided by 1.5, which is around, uh, 15.3. Well, it's exactly 15.33 meters per second and lastly we have P Q for which is equal to 80 minus 72 overturned Manus line 0.5 giving ghosts aid over zoo 0.5 meters about by 0.5 seconds, which is equal to 16 meters per second. So these are the slopes for each of the secret lines that we have now for B These asking us to approximate to say about. So we have to approximate how fast the object was going right before it hit the surface or when it hit the surface and the most accurate, um, calculation we have for this is this right here The secret line right here. And this slope right here represents a speed, as we see here, meters per second minutes per second per second meters per second. All of these were speeds. So toe answer B. We simply just use the most recent speed that we calculated which is peak you for. And we already found that to be 16 meters per second. That's it. Thank you.

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