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While searching for the answer regarding, why acceleration is needed to be constant for using the formula $(v_1+v_2)/2$ , I found many simple and easy proofs regarding this, here in this Physics.SE website, one of which is , enter image description here

But can anyone come up with a daily life simple explanation for understanding why acceleration is needed to be constant for using the formula $(v_1+v_2)/2$ , for a freshman student in physics like me.

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    $\begingroup$ Devil's advocate here: Actually that Phys.SE answer proves that constant acceleration is a sufficient assumption, not that it is necessary. $\endgroup$
    – Qmechanic ♦
    Sep 24, 2016 at 13:34

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Here's one way to think of it that might help.

If the acceleration is not constant you could have a case where something moves at a slow velocity for a long time and then accelerates briefly at the end of its motion to a higher velocity. Intuitively the average velocity should be closer to the initial slower velocity because it was travelling at that velocity for longer but the formula always puts halfway between initial and the final.

If there is a constant acceleration then half the time the velocity is slower than the average and half the time it's faster and the formula works.

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    $\begingroup$ Alright , i got your point. But, here as Devil wrote this is a "sufficient assumption , not that it is necessary", does that mean that there is some case possible where we can apply the above formula without the accleration being constant. $\endgroup$
    – Who I Am
    Sep 25, 2016 at 20:32
  • $\begingroup$ Sure, you could contrive some non constant acceleration that would make the equation work. $\endgroup$
    – M. Enns
    Sep 25, 2016 at 20:43
  • $\begingroup$ M.Enns , can you please come up with a simple example regarding this situation in which the formula works with non-uniform accleration too, because i am unable to think that high . $\endgroup$
    – Who I Am
    Sep 26, 2016 at 20:17
  • $\begingroup$ Sure - say you start at 2 m/s and accelerate steadily for 6 s to 4 m/s. Total distance is 18 m. Now say you start at 2 m/s accelerate for 2 s to 4 m/s, decelerate for 2 s back to 2 m/s and then accelerate for 2 s to 4 m/s. Total distance traveled once again is 18 m. It just a question of drawing some velocity vs. time graphs with equal areas. $\endgroup$
    – M. Enns
    Sep 26, 2016 at 21:07
  • $\begingroup$ This means , both way we get [(2+4)/2]m/s as the average velocity by using the formula (v1+v2)/2. Am i right ? $\endgroup$
    – Who I Am
    Sep 26, 2016 at 21:56
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A visually geometric answer is that the average velocity with respect to time over an interval is the area under a velocity-time curve (assuming rectillinear motion, of course), divided by the length of the time interval.

This area is only equal to the mean of the end velocities if the curve is a straight line, in which case the said mean is simply using the formula for the area of a trapezoid.

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