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Problem 17

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


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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

it goes. So this is Ah, question 20 in chapter two and section one. And it gives us the projectile motion off this object and asks us to estimate the slopes of these four seeking lines over here. Pick you one through, pick you for. And so to find the slope of a Sikh in line in between two points, we first have to calculate the rise and the run. And this is just the vertical distance and horizontal distance from one point to the other. And so the resulting slope is just equal to the rise over the run. So, for a I'm gonna start with pick you one. And, um, we're going from Q one over here to pee just like that. So we're gonna have to find the rise and then the run. We just clear that away. So the rise over here would be 80 minus 20. It's the height off p minus the height of Q one. And then, um, the, um distance along the x axis off p minus that of Q on. So that would be 10 minus five, which is 60/5, which gives those 12 meters per second. And I'm gonna calculate, um, and include the units. And we'll see why in would be we'll see why, Leader. So I'm gonna continue with pick. You two be maintained. Position so and the system 80. But que Tu is now at 38. And we have 10 minus, uh seven, which gives us 42/3 42 meters for every three seconds, which is 14 meters per second p Q three, we have 80 minus 57 over 10 minus 8.5, which gives us 23 meters divided by 1.5 seconds, which is 15.33 meters per second p Q. For is again 80 minus looks to be 72. Divide by turn, minus lam from five, which gives us eight meters for every 0.5 seconds, which is 16 meters per second. Simple. So these are all the slopes off the secret lines. And so for B, they want us to It asks about how fast was it going when it hit the surface? Now here they want us to estimate the speed of the object right before hit the surface. And, um, the most accurate representation of that we have is this speed of the object right before hit the surface. So now pee represents the surface of the moon and the speed that we have The closest speed we have to when the object hit the surface is P Q for. And as you can see, we already calculated that over here. So the answer is simply, just pick you for just 16 years per second. And now you see why? Um, I included the units for every calculation. Because all of these slopes every here represent speeds in this scenario, which fits for our answer for part B, and that's it.

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