# Acceleration of a free-falling body **URL:** \ **Category:** Factual Questions\ **Created:** [August 30, 2009, 6:29pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273 "2009-08-30T18:29:57Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [August 30, 2009, 6:29pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/1 "2009-08-30T18:29:57Z")
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?’): > [@](#): > > _v_ = 32_t_ > > This means that if a body is held motionless above the surface of the earth and is then dropped and allowed to fall freely, it would be moving at the end of one second at the rate of 32 feet per second; and at the end of two seconds, 64 feet per second; at the end of three seconds, 96 feet per second; and so on. So far, so good. Then it goes on to say: > [@](#): > > The equation worked out by Galileo for the distance, _d_, covered by a freely falling body was > > _d_ = _gt_[sup]2[/sup]/2 > > Since _g_ is equal to 32, the equation works out to > > _d_ = 16_t_[sup]2[/sup] > > This means that, after 1 second, a freely falling body covers a distance of (16)(1)(1) or 16 feet; after 2 seconds, it covers a distance of (16)(2)(2) or 64 feet; after 7 seconds, a distance of (16)(7)(7) or 784 feet, and so on. OK. So at any given time, the velocity of a falling body is time multiplied by 32, or _v_ = 32_t_. The distance covered by the falling object is _d_ = _gt_[sup]2[/sup]/2 or _d_ = 16_t_[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? 16_t_ 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 32_t_[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. Thanks in advance. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 30, 2009, 6:58pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/2 "2009-08-30T18:58:19Z")
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. ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [August 30, 2009, 7:08pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/3 "2009-08-30T19:08:37Z")
> [@Indistinguishable](#): > > 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. ---
**Author:** ![Simplicio](https://avatars.discourse-cdn.com/v4/letter/s/c37758/32.png) [@Simplicio](https://boards.straightdope.com/u/Simplicio)\ **Post date:** [August 30, 2009, 7:10pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/4 "2009-08-30T19:10:05Z")
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 ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [August 30, 2009, 7:33pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/5 "2009-08-30T19:33:19Z")
**Simplicio** : As I said, it’s been a very long time… 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 16_t_[sup]2[/sup], which, as I said, makes perfect sense, and the rate of acceleration is 32_t_[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. ---
**Author:** ![The\_Hamster\_King](https://avatars.discourse-cdn.com/v4/letter/t/8edcca/32.png) [@The\_Hamster\_King](https://boards.straightdope.com/u/The_Hamster_King)\ **Post date:** [August 30, 2009, 7:34pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/6 "2009-08-30T19:34:55Z")
> [@Johnny\_L.A](#): > > If I substitute 32_t_[sup]2[/sup] for _r_ in the last equation … . How are you doing that? There is no r in the last equation! ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [August 30, 2009, 7:44pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/7 "2009-08-30T19:44:49Z")
> [@The\_Hamster\_King](#): > > How are you doing that? There is no r in the last equation! Sorry. I was thinking _rt_ = _d_. But of course, that was wrong-thinking; since the rate is constant, and in 32 ft/sec/sec the rate is changing. ---
**Author:** ![The\_Hamster\_King](https://avatars.discourse-cdn.com/v4/letter/t/8edcca/32.png) [@The\_Hamster\_King](https://boards.straightdope.com/u/The_Hamster_King)\ **Post date:** [August 30, 2009, 7:45pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/8 "2009-08-30T19:45:24Z")
> [@Johnny\_L.A](#): > > So my question is basically where to use 32 and where to use 16. Never use 16. Always use 32. 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. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 30, 2009, 7:53pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/9 "2009-08-30T19:53:26Z")
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. ---
**Author:** ![johnpost](https://avatars.discourse-cdn.com/v4/letter/j/f17d59/32.png) [@johnpost](https://boards.straightdope.com/u/johnpost)\ **Post date:** [August 30, 2009, 8:05pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/10 "2009-08-30T20:05:55Z")
> [@Johnny\_L.A](#): > > 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 32_t_[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. > > Thanks in advance. acceleration is a rate of change of velocity. the value of acceleration due to gravity on earth is about 32 ft / s[sup]2[/sup] it seems confusing to refer to a rate of acceleration (which itself is the rate of a rate). ---
**Author:** ![Stealth\_Potato](https://avatars.discourse-cdn.com/v4/letter/s/d78d45/32.png) [@Stealth\_Potato](https://boards.straightdope.com/u/Stealth_Potato)\ **Post date:** [August 30, 2009, 8:39pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/11 "2009-08-30T20:39:25Z")
> [@Johnny\_L.A](#): > > Where I’m confused is that distance is 16_t_[sup]2[/sup], which, as I said, makes perfect sense, and the rate of acceleration is 32_t_[sup]2[/sup]. Not sure if this is just a typo or not, but 32_t_[sup]2[/sup] isn’t a rate of acceleration; if the units of _t_ are seconds, then 32_t_[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]. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 30, 2009, 8:47pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/12 "2009-08-30T20:47:05Z")
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. ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [August 30, 2009, 9:00pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/13 "2009-08-30T21:00:33Z")
> [@Chronos](#): > > Note, by the way, that the acceleration can’t be just “32”. It has to have units. Shorthand for the whole thing. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [August 31, 2009, 2:18am UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/14 "2009-08-31T02:18:30Z")
> [@Johnny\_L.A](#): > > Now, I’ve always heard that the rate of acceleration is 32 ft/second[sup]2[/sup]. It’s often expressed that way, but that’s sloppy usage. Acceleration is the rate at which velocity changes (IOW velocity/time). So to be accurate, omit “rate of” in the above statement, and simply say that acceleration is 32 ft/second[sup]2[/sup]. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 31, 2009, 5:59am UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/15 "2009-08-31T05:59:01Z")
To my ears, there’s nothing wrong with the expression “rate of acceleration”, even if it is slightly redundant. ---
**Author:** ![groman](https://avatars.discourse-cdn.com/v4/letter/g/73ab20/32.png) [@groman](https://boards.straightdope.com/u/groman)\ **Post date:** [August 31, 2009, 10:41am UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/16 "2009-08-31T10:41:34Z")
> [@Indistinguishable](#): > > To my ears, there’s nothing wrong with the expression “rate of acceleration”, even if it is slightly redundant. How do you differentiate acceleration A from the rate of acceleration dA/dt? ---
**Author:** ![Toolie](https://avatars.discourse-cdn.com/v4/letter/t/41988e/32.png) [@Toolie](https://boards.straightdope.com/u/Toolie)\ **Post date:** [August 31, 2009, 1:04pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/17 "2009-08-31T13:04:30Z")
> [@Chronos](#): > > If you’re using lightyears and years for your units, then it’s about 1 ly/yr[sup]2[/sup]. The amount of awesome that this is correct to within about 3% cannot be overstated. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [August 31, 2009, 1:51pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/18 "2009-08-31T13:51:58Z")
> [@Johnny\_L.A](#): > > Shorthand for the whole thing. 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). ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 31, 2009, 1:55pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/19 "2009-08-31T13:55:55Z")
> [@groman](#): > > How do you differentiate acceleration A from the rate of acceleration dA/dt? 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”. ---
**Author:** ![Really\_Not\_All\_That\_Bright](https://avatars.discourse-cdn.com/v4/letter/r/e8c25b/32.png) [@Really\_Not\_All\_That\_Bright](https://boards.straightdope.com/u/Really_Not_All_That_Bright)\ **Post date:** [August 31, 2009, 2:21pm UTC](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273/20 "2009-08-31T14:21:04Z")
Of course, this all assumes that your object doesn’t reach terminal velocity within 784 feet. [Next page](https://boards.straightdope.com/t/acceleration-of-a-free-falling-body/508273.md?page=2)