# More odd ways of thinking about acceleration

I’ve been thinking a lot about the sequence of odd numbers and its relationship to accelerated motion.

1, 3, 5, 7, 9, 11, 13…

In a previous post, I came to this sequence by discussing a special kind of motion that I defined as always covering 3x the distance in the second half of a trip than in the first. You can check that the odd number sequence meets this criteria.

Moving up the sequence, each number merely represents the Δx covered in successive intervals of time. Because they are equal time intervals, they also represent the average velocity over successive intervals. The sequence shows that the average velocity is increasing by 2 chunks each second. This of course means, that the ball has a constant acceleration of “+2”

You might think that we’d have to craft a whole new sequence for different accelerated motions. But you don’t. If we want a sequence with twice as much acceleration, we just multiply each number in the sequence by two. If you want to cut the acceleration in half, just half every number in the sequence.

You can also use this sequence to describe motions not starting from rest. For example, if you want something that starts with some initial velocity and speeds up, you just start somewhere else in the sequence besides the beginning. If you want something slowing down, you just move backwards in the sequence.

I’m not saying this is anything new. I’m just saying it’s interesting to think that sequence 1, 3, 5, 7, 9… is a representation for constant acceleration, and indeed any motion involving constant acceleration.

**Does this way of thinking help us to solve problems?**

Let’s say you want to answer the question, what is the acceleration of a ball that rolls down a 16ft ramp in 4 seconds?

All you do is try to come up with the sequence of 4 numbers that gets you to 16 ft. This one is easy: 1 *(1 + 3 +5 +7 ) = 16 ft. Since the series steps in increments of 2, the speed is increasing by +2 ft each second. As a result, the acceleration is 2 ft/s each second.

I personally thought it would be hard to come up with the sequence for 200 cm in 10 seconds, but it’s not. The answer is

2 * (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19) = 2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 + 34 + 38. This clearly has an accelerated of 4 ft/s each second. It’s easy to check my sequence adds to 200 by adding in groups of 40, which is the insight you need to make producing these sequences so easily. Getting good at this takes some practice, and it is still terribly difficult for many situations. But the idea is simple: **The numbers you add up must be the odd number sequence or a multiple of it. **

OK. Where to next? One way of moving beyond writing down a million sequences for every new problem is to benefit from the fact that a sequence of odd numbers is always a square. For example,

1 + 3 = 4

1 + 3 + 5 = 9

1 + 3 + 5 + 7 = 16

…

In general if you are summing to the Nth odd number, then Σ (2n -1) = N² … this is of course the same as saying that accelerated motion goes like t².

Aside: You can prove this equality by showing that difference between two successive squares is always an odd number… n² – (n-1)² = n² – (n²-2n +1) = 2n-1

This tells us that our original sequence of odd numbers, which we said shows an acceleration of “+2”, goes like Σ(Δx) = n² , where n is the final place in the sequence. In other words, if you go for 10 chunks of time, you get to 100 chunks of distance.

This allows you to some really nice estimations: Let’s say you are trying to figure out what acceleration is for a ball covering 372 ft in 19 seconds (starting from rest).

Let’s use our rule of squares: 19² = 361ft, which is the distance that an object accelerating at 2 ft/s/s would get after 19 seconds. Given that we landed at 372, the acceleration must be a little big bigger than 2ft/s/s? How much bigger? Well, 372/361 = 1.03… or three percent further than you’d expect. Thus the acceleration should also be 3% greater than expected, or 2.06 ft/s/s

If you want to write the sequence down, it’s just 1.03 * (1+3+ … + 35 + 37) = 1.03 (19)² = (2.06 /2) (19)²

**Why worry about this?**

I don’t think you have to. But I do think that reasoning from well-understood special cases is an important skill. The a =2 case is really compelling and interesting to think about because

The successive distances are simply the sequence of odd numbers

The successive positions are simply the sequence of square numbers

**Understanding how this special case relates to other cases is also interesting**

(1) Different magnitudes of acceleration are merely multiples of the odd number sequence

(2) Starting with some initial velocity is merely starting mid-sequence

(3) Slowing down is running the sequence backwards

I know I have a different way of thinking about why accelerated motion goes like time squared, and I have some news tools make it really easy to estimate accelerations, positions, etc

Why are the odd numbers better than the counting numbers?

Most of what you said is true of 1,2,3,4,5 as well. It’s sum grows as n squared. It has constant growth, etc.

I missed this somehow… I think the natural numbers better correspond to m (m+1)…right?

Beginning with m=0, we get 0, 2, 6, 12, 20… and looking at 1st differences we get (2, 4, 6, 8) which is 2 (1,2,3,4,…) which is the natural number sequence.