I picked up Realm Of Algebra by Isaac Asimov based on a comment in another thread. Even though I already knew the math in it, I thought it would be an interesting read. And it was. But here’s something near the end (the chapter is basically ‘What good is knowing this stuff?’):
So far, so good. Then it goes on to say:
OK. So at any given time, the velocity of a falling body is time multiplied by 32, or v = 32t. The distance covered by the falling object is d = gt[sup]2[/sup]/2 or d = 16t[sup]2[/sup]. This makes sense, because if an object is falling at a velocity of 32 ft/second at the end of one second, it can’t have fallen 32 feet in one second; it took a second to reach that speed. Rate times time equals distance. So if an object falls 784 feet, how much time did it take?
16t x t = 784; or t[sup]2[/sup] = 49; or t = 7 seconds.
Just as Asimov wrote. Now, I’ve always heard that the rate of acceleration is 32 ft/second[sup]2[/sup]. Unfortunately, it’s been far too long since I’ve played with this stuff; and Asimov, while he wrote about velocity at a given time, and of distance, didn’t mention rate. If I substitute 32t[sup]2[/sup] for r in the last equation, I come up with (32)(49) = 784; which it isn’t. But that’s because the rate is the rate at a given time and not the rate of acceleration. So please remind me how I can use the rate of acceleration of 32 feet per second per second to solve a problem.
The distance covered over a timespan is equal to the amount of time * the average speed (with respect to time). So if you start from a velocity V[sub]0[/sub] and accelerate for time t at a constant acceleration of A, you end up at a velocity of V[sub]0[/sub] + At, and your average velocity (with respect to time) over that interval was V[sub]0[/sub] + At/2 (because your velocity increased linearly over that span). Thus, the distance you covered during that time was V[sub]0[/sub]t + At^2/2. In particular, in the case where you started from rest, the distance you cover is At^2/2. Which shows you how to correctly derive your formula “16t^2” from your specific rate of acceleration “32”.
This is just a special case of the idea of “integration”, which is illuminated in greater detail by the methods of calculus.
It’s been like three decades since I’ve done any calculus. (I may have to sign up for a refresher.) Can you give an example of a problem that uses 32 ft/sec[sup]2[/sup]? If I can see it ‘used in a sentence’, as it were (i.e., a practical example), then I think I’d have a better grasp of it.
I’m having a little trouble figuring out what your asking from your post, but I think your confusion is arising from the “seconds[sup]2][/sup]” in the statement acceleration=32 ft/second[sup]2[/sup]
Distance per time squared doesn’t mean that you square time to get the acceleration, it’s just the units of acceleration. So 32ft/second[sup]2[/sup] is a constant acceleration, not one that changes with the square of the amount of time that’s elapsed.
This is analogous to velocity having units of meters per second, a object traveling 6 meters a second for six seconds isn’t going 6*6=36 meters per second, it’s velocity is constant and so it’s still traveling at 6 meters a second.
Hope that helps, and apologies if I misunderstood your problem
If I drop an object (in Earth’s gravity, in a vacuum, etc.) it’s speed will be 32 ft/sec after one second. It’s speed will be 64 ft/sec after two seconds. At the end of four seconds it’s speed will be 128 ft/sec. This is what I mean: that speed equals 32 times the time. So given a ‘prefect system’ where there’s no air resistance, nothing to run into, and gravity remains constant regardless of the distance from the mass, and ignoring relativistic aspects, and so on, an objects speed will inrease over time. The rate of acceleration stays the same, but the velocity increases.
Where I’m confused is that distance is 16t[sup]2[/sup], which, as I said, makes perfect sense, and the rate of acceleration is 32t[sup]2[/sup]. This also makes sense, since we’re talking about a change in velocity and not the velocity itself or time. So my question is basically where to use 32 and where to use 16. The question isn’t clear because it’s been too long since I’ve thought about such things and don’t remember enough to pose a question in a manner that makes sense to people who haven’t forgotten. If you know what I mean.
The equation for distance traveled is d = 32 * t^2 / 2. Just because you can simplify it by dividing the 32 by the 2 in advance, that doesn’t mean that the 16 means anything. The rate of acceleration is still 32 ft/s^2.
Sure, the 16 means something. 16 feet/sec^2 is the rate of change of your average speed (that is, averaged over time), just as 32 feet/sec^2 is the rate of change of your actual speed.
I think if you understand where the formulas come from, you’ll be better off. I tried to give an explanation of this in post #2, but if there’s something you’d like more clarification on, by all means, ask.
Not sure if this is just a typo or not, but 32t[sup]2[/sup] isn’t a rate of acceleration; if the units of t are seconds, then 32t[sup]2[/sup] is just a scalar value whose magnitude is (32 * t * t) and whose units are “square seconds.” (Analogous to “square feet.”)
The “rate” of acceleration is 32 feet / second[sup]2[/sup].
Note, by the way, that the acceleration can’t be just “32”. It has to have units. If you’re using feet and seconds as your units, then it’s about 32 feet/s[sup]2[/sup]. If you’re using meter and seconds as your units, then it’s about 9.8 m/s[sup]2[/sup]. If you’re using lightyears and years for your units, then it’s about 1 ly/yr[sup]2[/sup]. The units are an essential part of the quantity.
However, it’s apparently the lack of units that’s confusing you.
Acceleration is 32 feet per second per second, or 32 ft/sec[sup]2[/sup]. The “ft/sec[sup]2[/sup].” are the units.
That’s not the same thing as 32t[sup]2[/sup]. 32t[sup]2[/sup] has units of either “seconds squared” (is the 32 is unitless) or “ft” (if the 32 is an acceleration).
I wouldn’t call dA/dt “the rate of acceleration” anymore than I would call acceleration itself “the rate of speed”. I would call dA/dt “the rate of change of acceleration”.