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I understand that the concept of an average of a data list means finding a certain value 'x', which ensures that the sum of the deviations of the numbers on the left of 'x' and on the right of 'x' yields the value 0. Assume we are given the values 2,6,9 and 12. I label these values from left to right in the sequence I have provided as: a,b,c and d respectively. The definition of finding these values' average (let's call this average value 'x') in mathematical form is:

$${(x - a) + (x - b) = -(x - c) - (x - d)}$$

So this equation can be reduced to:

$$\Large x = \frac{(a + b + c + d)}{4}$$

This will provide you with the value for the average of any number of values. Similarly there is a formula in physics that calculates the average velocity from a graph by dividing the sum of the initial and final velocities by 2.

$$\Large\frac{v_f + v_i}{2} = \bar{v}$$

There is also another formula which calculates the average velocity from a graph by calculating the slope of the line through two points on the graph. This is done by dividing the change in position by the change in time:

$$\Large \frac{x_f - x_i}{t_f - t_i} = \bar{v}$$

I can understand the third equation from the top of the page, as I can understand this calculation of average velocity in terms of what I explained in the first two equations from the top of the page. I am unable to reconcile the calculation of the average of velocities as seen in the last equation with the method as listed above in the first two equations. Can someone help me solve this dilemma?

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  • $\begingroup$ One thing, the first two equations tell a different tale. They are written for discrete data, while the velocity of an object w.r.t time is a continuous data. You can derive last equation from the generalization of first two. Whereas, the third equation is a special case of the last, when the acceleration is constant. Hope that helps! $\endgroup$ – Cheeku Mar 4 '13 at 12:07
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  1. The temporally weighted average velocity $\bar{v}$ is defined as $$\tag{1} \bar{v} ~=~ \frac{\int_{t_i}^{t_f}\! dt~ v(t) }{\int_{t_i}^{t_f}\! dt} ,$$ where
    $$\tag{2} v(t) ~:=~ \frac{dx(t)}{dt}$$ is the instantaneous velocity at instant $t\in[t_i,t_f]$. By use of elementary integration and differentiation calculus, equation (1) can also be written as $$\tag{3} \bar{v} ~=~ \frac{\Delta x }{\Delta t } ,$$ where $\Delta x:=x_f-x_i$ and $\Delta t:=t_f-t_i$.

  2. If the instantaneous acceleration $$\tag{4} a(t) ~:=~ \frac{dv(t)}{dt} $$ is constant in time, then it is possible to prove that the equation (1) can be rewritten as $$\tag{5} \bar{v} ~=~ \frac{v_f+v_i}{2} .$$ Can you see how the proof goes?

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I think your problem is that your two calculations for the average velocity are different and will generally return different values.

For example, suppose I travel 100 metres in 100 seconds the your third equation gives my average speed as 1 m/sec, and I guess most of us will find this a reasonable way to calculate the average. However suppose I told you that I travelled the first 99 metres at 0.991 m/sec and the last metre at 9.91 m/sec i.e. $v_i$ = 0.991 and $v_f$ = 9.91. If you work out my travel time it still adds up to 100 seconds, so I still travelled 100 metres in 100 seconds. However your second equation would give the average velocity as 5.45 m/sec. The problem with the second equation is that it gives the same weight to the two velocities even though I was travelling at the faster velocity for only a very short time.

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  • $\begingroup$ But I was reading in a physics book about the derivation of the equations of motion, where the author equates the first equation for calculating average velocity to the second equation for calculating average velocity. $\endgroup$ – Ram Sidharth Mar 4 '13 at 12:22
  • $\begingroup$ The second and third equations will give the same result if the acceleration is constant, but only if the acceleration is constant. $\endgroup$ – John Rennie Mar 4 '13 at 15:32

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