Understanding why a term is squared

[MrCuriosity]

I had a problem that I understand how to do the math part, I just don't understand a certain part of it. Here is the problem: If a car can go from 0 to 24.6 meters per second in 12 seconds, what is its acceleration? The answer is 2.05 meters per second. But what I don't understand is why is it supposed to be termed 2.05 meters per second squared? How is it squared? I understand that for every second of travel the vehicle goes an additional 2.05 meters, but at no point in this equation is there any squared operation. Can anyone explain this for a tenth grade mind to understand?

 

[srmichael]

I had a problem that I understand how to do the math part, I just don't understand a certain part of it. Here is the problem: If a car can go from 0 to 24.6 meters per second in 12 seconds, what is its acceleration? The answer is 2.05 meters per second. But what I don't understand is why is it supposed to be termed 2.05 meters per second squared? How is it squared? I understand that for every second of travel the vehicle goes an additional 2.05 meters, but at no point in this equation is there any squared operation. Can anyone explain this for a tenth grade mind to understand?

If you look at this including the units I think it will make sense to you.

\(\displaystyle \dfrac{24.6m/sec}{12sec} = \dfrac{24.6m}{1sec} \cdot \dfrac{1}{12sec} = \dfrac{24.6m}{12sec^2} = 2.05 m/sec^2\)

 

[HallsofIvy]

"Speed" is "change in distance divided by time" so has units of "miles per hour" or "meters per second".

"Acceleration" is "change in speed divided by time" so the units are "miles per hour" or "meters per second" divided by time: seconds again. So the units would be "(miles per hour) per hour" or "(meters per second) per second".
Those would be represented as "(miles/hour)/hour= \(\displaystyle miles/hour^2\)" or "(m/s)/s= \(\displaystyle m/s^2\)".

Of course, none of those are numbers so you are not really "squaring" "hours" or "seconds". It is a useful notation that reminds you that acceleration involves dividing distance by time twice.

 

[MrCuriosity]

Ok, I think I understand what you are each saying, but my mind is making me wonder what the squared is still about. I am I correct in thinking that if you square the 2.05, that is the averaged distance that you are traveling for each second of the 12 seconds? So after 12 seconds you've traveled 50.43 meters? That makes more sense than if you traveled 2.05 meters in the first second, and then square that to tell you how far you've went by second number 2 (4.2), and then continue to square each subsequent value until 12. The number gets way too high by then. I tried it in Excel and it couldn't calculate it after 10 seconds, which means I've probably traveled to the moon by then.

Time (sec)	0	1	2	3	4	5	6	7	8	9	10	11	12
Speed (m/sec)	0	2.05	4.1	6.15	8.2	10.25	12.3	14.35	16.4	18.45	20.5	22.55	24.6
Distance (m)	0	2.05	4.2025	17.66100625	311.9111418	97288.56036	9465063976	8.95874E+19	8.02591E+39	6.44152E+79	4.1493E+159	#NUM!	#NUM!

Sorry, I couldn't get the lines to show up. The first row is 0 to 12 seconds. Second row is the Speed in m/sec. Third row is Distance in meters.

 

[MrCuriosity]

Now that I look at this a little closer, I think the 50.43 is too short of a distance. If I'm capable of traveling at 2.05m/s after the first second, and then I'm capable of traveling at 4.1m/s by the second second and so on, I should be able to add each of those up and by second 12 I should have traveled at least 116 meters.

I'm beginning to think this over too much. Is there a better way to solve this? Am I getting into levels of math that are much higher than what I'm in, or will I eventually see this in algebra?

 

[JeffM]

Now that I look at this a little closer, I think the 50.43 is too short of a distance. If I'm capable of traveling at 2.05m/s after the first second, and then I'm capable of traveling at 4.1m/s by the second second and so on, I should be able to add each of those up and by second 12 I should have traveled at least 116 meters.

I'm beginning to think this over too much. Is there a better way to solve this? Am I getting into levels of math that are much higher than what I'm in, or will I eventually see this in algebra?

\(\displaystyle Let\ distance\ traveled\ during\ period = d.\)

\(\displaystyle Let\ time\ that\ period\ lasts = t.\)

\(\displaystyle Let\ average\ velocity\ during\ period = v = \dfrac{d}{t}.\) This is all straightforward. You with me to here?

If something is accelerating (or decelerating) during a period, the velocity is changing during the period. So the average velocity is just that, an average of different velocities. Still clear?

We are not going to consider the complicated case where acceleration is changing during the period. That takes calculus. We shall assume that acceleration is constant and positive.

We are also going to keep things simple by assuming that the velocity at the start of the period is zero.

\(\displaystyle Let\ the\ final\ velocity\ at\ the\ end\ of\ the\ period = f > 0.\)

The technical definition of acceleration is the change in velocity during the period divided by the time of the period.

The change in velocity is f - 0 = f. The time of the period = t. So

\(\displaystyle Let\ acceleration = a = \dfrac{f}{t}.\) How you holding up? That also means \(\displaystyle f = at.\)

Obviously f will be larger than v, because v is an average for the period and f is the highest velocity achieved. Still with me?

In fact, \(\displaystyle v = \dfrac{f + 0}{2} \implies f = 2v = 2 * \dfrac{d}{t} = \dfrac{2d}{t} \implies a = \dfrac{f}{t} = f * \dfrac{1}{t} = \dfrac{2d}{t} * \dfrac{1}{t} = \dfrac{2d}{t^2}.\)

Now do you see where the square comes from. We divided by t once to get velocity. We divided by t a second time to get acceleration.

The formula for distance given constant acceleration from an initial velocity of 0 is \(\displaystyle d = \dfrac{1}{2} * at^2.\)

Where in the world did that crazy looking formula come from?

\(\displaystyle d = vt.\)

\(\displaystyle But\ f = 2v \implies d = vt = \dfrac{1}{2} * ft = \dfrac{1}{2} * at * t = \dfrac{1}{2} at^2.\)

So let's look at your example.

\(\displaystyle f = \dfrac{24.6\ m}{sec}.\)

\(\displaystyle So\ f = 2v \implies v = \dfrac{12.3\ m}{sec.}\)

\(\displaystyle And\ a = \dfrac{f}{t} = \dfrac{24.6\ m\ per\ sec}{12\ sec}= \dfrac{24.6\ m}{sec} * \dfrac{1}{12\ sec} = \dfrac{2.05\ m}{sec^2}.\)

\(\displaystyle d = \dfrac{1}{2} * at^2 = \dfrac{1}{2} * \dfrac{2.05\ m}{sec^2} * (12\ sec)^2 = 1.025 * 144\ m = 147.6\ m.\)

We get the same answer by looking at \(\displaystyle d = vt = \dfrac{12.3\ m}{sec} * 12\ sec = 147.6\ m.\)

 

[wjm11]

I had a problem that I understand how to do the math part, I just don't understand a certain part of it. Here is the problem: If a car can go from 0 to 24.6 meters per second in 12 seconds, what is its acceleration? The answer is 2.05 meters per second. But what I don't understand is why is it supposed to be termed 2.05 meters per second squared? How is it squared? I understand that for every second of travel the vehicle goes an additional 2.05 meters, but at no point in this equation is there any squared operation. Can anyone explain this for a tenth grade mind to understand?

...Ok, I think I understand what you are each saying, but my mind is making me wonder what the squared is still about. I am I correct in thinking that if you square the 2.05, that is the averaged distance that you are traveling for each second of the 12 seconds? So after 12 seconds you've traveled 50.43 meters?

Hello, MrCuriosity,

Yours is a great question and a common point of confusion among students. The answer: Only the units are squared (the seconds) -- NOT the number! Do not apply the "squaring" to the "2.05".

***

As an aside, to calculate the distance traveled, we have two different cases, each with its own formula.

When an object moves at constant velocity, distance equals velocity times time: d = vt (You've probably known that formula for a long time.)

When an object starts at zero velocity and is under constant (steady) acceleration, distance equals one-half times acceleration times time squared: d = (1/2)(a)(t^2)

Let's examine the "t^2" term as it applies to this problem. Since t = 12s, t^2 = (12s)^2. Notice that the exponent is outside the parentheses around the "12s". In this case, both the "12" and the "s" will be squared: (12s)^2 = 144 s^2.

So the distance traveled is:

d = (1/2)(a)(t^2) = (1/2)(2.05 m/(s^2))(144 s^2) = 147.6 m

Notice that the two "s^2" terms cancel each other out, because one is on top (in the numerator) and the other is on the bottom (in the denominator). That leaves just the "m" (for meters) in our answer, which makes sense, because we were calculating a distance. The units must work out correctly; they are very important in these calculations.

Hope that helps.