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(II) For an object falling freely from rest, show that the distance traveled during each successive second increases in the ratio of successive odd integers $(1,3,5,$ etc. $)$ . (This was first shown by Galileo.) See Figs. 26 and $29 .$ FIGURE 26 Multiflash photograph of a falling apple, at equal time intervals. The apple falls farther during each successive interval, which means it is accelerating.

$1,3,5,7, \ldots$

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

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So we have an object, uh, falling freely. Um, And in the diagram, it's just a ball dropped. And so we can model this, uh, motion by can America equations by saying delta Y is ah, are Notre Velocity Times T plus 1/2 A T squared. And now V zero is zero because we're dropping from rest and a is minus g uh, where something. We're on earth. And we want to show something interesting that the distance traveled every second. Ah, is increasing by odd integers. And it seems, um, a little weird. Maybe not intuitive, But let's see what we get when we analyze this soap. Ah, Delta. Why? Like I said, B zero is zero. So we get minus 1/2 g t squared and Delta Y is just wife T minus our initial position. In our initial position, we'll just call zero to make it easier. So we get y t equals minus 1/2 to t squared. Now we just want to know the distance travelled, not necessarily the displacement. And when the when the ball is falling, it's going down, which is what's making this negative. So I'm just gonna call it positive uh, because we just like I said, we just care about the distance, not the total displacement. So I'll say, Why have tea? It's 1/2 a t squared and now we have, ah, position as a function of time. So why don't I just plug in time in successive every second and see what we get out of that? So wife one is 1/2 g times one squared, which is just 1/2 g. And to make it easier, I'm gonna call this K and K is the distance it travels in the 1st 2nd And if we plug in to we get 1/2 G times two squared equals. See, this is four. I'm keeping us four halves g because we can see this is four K. I'll do one more wife three. We get three squared, just nine halves G just nine K. Now, if you want to find the distance for every second, we can just subtract. So why I have to? Minus wife one. It's four k minus k. This three k and I guess we can also say why have one minus y zero is K. His wife's here was zero, and for the last one. Wife three minus wife too. Okay, it is nine K minus four K. It's five K. So we can kind of see the pattern coming out here. We have one K three K five K. Now it looks like it's gonna be auditors, but if we want to prove it explicitly Ah, we can do this for any Why end? It's going to be k times and squared. And why have n plus one the next second que times and plus one squared and the distance travelled between wife and plus one and white and is K times and plus one squared minus K times end squared. And so if we expand Ah, this we get k times and squared plus two n plus one and this is and squared. I mean, now we can factor a k So we have k times and squared plus two n plus one minus and squared. And this comes out to K times Ah, two and plus one. So for every second we increase, we get to n plus one k. And if you're familiar with this to end plus one is how we represent an odd integer cause for any input for any integer input, you get to end which makes it automatically even. And then plus one makes it odd. So let's check. Ah, we have Why zero We get K times two times zero plus one, which is one que ah, why of one again, This is the same thing we were doing in the first part. Except now we just have unaudited your instead of a square. This is gonna be three k. Uh, The next one is five k. You can check, keep going and you'll see that for every increase in a second, the distance traveled. Ah is increasing in successive odd integers And this because of the square relationship between distance and time, when we have a constant acceleration and free fall.

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