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:
<br /> \\<br /> v(t) = 8 + 5t + \frac{3t^2}{2} \\<br /> a(t) = 5 + 3t \\<br /> <br /> v(t=5) = 70.5 m/s;v(t=0)= 8 m/s \\<br /> a(t=5) = 20 m/s^2;a(t=0) = 5 m/s^2\\ \\ \\<br /> <br /> a_m(\Delta t) = \frac{v(t=5) - v(t=0)}{5-0} = 31.25m/s^2 \\ \\<br /> a_m&#039;(\Delta t) = \frac{20+5}{2} = 12.5 m/s^2<br />

 

[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'.