Why Is Medium Acceleration Calculated Using Change in Velocity Over Time?

[ShizukaSm]

This I something I always wondered, but never could get my finger into:

$a_m = \frac{v_2-v_1}{t_2-t_1}$


Alright, that's medium acceleration, but why isn't it:

$a_m = \frac{a_2-a_1}{t_2-t_1}$

I mean, it does make perfect sense in my mind. Is it just because it was defined that way?

 

[HallsofIvy]

I have no clue why that would "make perfect sense"! What are "$a_2$" and "$a_1$"? Since you are using "a" to represent acceleration, I would assume that they are accelerations at some points but if your formula for acceleration involves knowing them, how would you find those accelerations to begin with?

Also, the units are wrong. In the MKS system acceleration has units of $m/s^2$. Subtracting two accelerations would give a numerator still in $m/s^2$ but then dividing by time, in seconds would give $m/s^3$ which cannot be equal to acceleration.

Acceleration is defined the "rate of change of speed"- $v_2- v_1$ is a "change in speed" and dividing by time tells how fast that change occured.

By the way, in English, the phrase is "average acceleration", not "medium. And you may be thinking of the average as "$(a_1- a_2)/2$". Now, that would give the average acceleration over an interval but you typically do not know the acceleration at two different times.

 

[ShizukaSm]

Yes. I apologize for my messiness, thanks for correcting my english too, I often write things that sound weird, it's with those corrections that I learn.

Let me be more specific. $a_2$ meant instantaneous acceleration at point two, $a_1$ at point one.

Indeed I meant $\frac{(a_2+a_2)}{2}$(arithmetic mean) , and it doesn't always give me the average acceleration. That was what I meant by my question, sometimes it gives different answers.

Let me give you an example:
$$\\ v(t) = 8 + 5t + \frac{3t^2}{2} \\ a(t) = 5 + 3t \\ v(t=5) = 70.5 m/s;v(t=0)= 8 m/s \\ a(t=5) = 20 m/s^2;a(t=0) = 5 m/s^2\\ \\ \\ a_m(\Delta t) = \frac{v(t=5) - v(t=0)}{5-0} = 31.25m/s^2 \\ \\ a_m'(\Delta t) = \frac{20+5}{2} = 12.5 m/s^2$$

 

[Chestermiller]

You calculated a_m incorrectly. You divided by 2 rather than 5. 62.5/5 = 12.5.

In general, you should not expect the average acceleration to be equal to the arithmetic average of the accelerations at the at the beginning and end of the interval. In this particular case, since the acceleration is varying linearly with time, a_m = a_m'.

 

[Nickie]

I can understand your confusion and curiosity about the formula for medium acceleration. The reason why we use the change in velocity over time (v2-v1)/(t2-t1) instead of the change in acceleration over time (a2-a1)/(t2-t1) is because acceleration is a measure of how quickly velocity changes, not how quickly acceleration changes. In other words, acceleration is the rate of change of velocity, not the rate of change of acceleration.

Think of it this way, when we say something has a medium acceleration, we are saying that its velocity is changing at a medium rate. So, it makes more sense to use the change in velocity over time to measure the medium acceleration. On the other hand, if we were talking about how quickly acceleration itself is changing, then we would use the change in acceleration over time.

Additionally, the units for acceleration are meters per second squared (m/s^2) which represents the change in velocity over time. So, using the change in velocity over time as the formula for medium acceleration aligns with the units we use to measure it.

Lastly, as scientists, we use formulas and equations to represent and describe the physical world. While it may seem arbitrary at times, these formulas are based on careful observations and experiments, and have been tested and proven to accurately represent the behavior of objects and systems in the real world. So while it may seem like it was just defined that way, there is sound reasoning and evidence behind the formula for medium acceleration.

I hope this explanation helps clear up your confusion and shows the importance of using precise and accurate formulas in the field of science. Keep questioning and seeking answers, that's what drives scientific progress!