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I am having trouble grasping how the average velocity of a particle between an initial position and final position equals to the arithmetic mean of the initial velocity and the final velocity when the acceleration is constant.

My textbook offers me the following explanation:

Because the velocity at constant acceleration varies linearly in time according to: $$v_{xf}=v_{xi}+a_{x}t,$$ we can express the average velocity in any time interval as the arithmetic mean of the initial and final velocity.

$$v_{x, average}=\frac{v_{xf}+v_{xi}}{2}$$ (for constant $a_{x}$)

I tried graphing a theoretical position-time, velocity-time graph to understand this but I still cannot see why it is the arithmetic mean.

Could someone please explain this to me?

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marked as duplicate by David Z Mar 13 '16 at 8:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$a_x \Delta t = \Delta v_x = v_{xf} - v_{xi}$

$\Delta x = v_{x,average}\Delta t = v_{xi}\Delta t + \dfrac{1}{2}a_x (\Delta t)^2$

$\Rightarrow v_{x,average} = v_{xi} + \dfrac{1}{2}a_x \Delta t = v_{xi} + \dfrac{1}{2}(v_{xf} - v_{xi}) = \dfrac{v_{xf}+ v_{xi}}{2}$


Is there a geometric interpretation or does it just work out mathematically?

enter image description here

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  • $\begingroup$ Sorry it's still unclear to me how you went from first to second line. $\endgroup$ – Bobby Jul 16 '13 at 20:55
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    $\begingroup$ The 1st and 2nd lines are independent. The 3rd line follows from the 1st and 2nd lines. What is unclear in the 2nd line? The average velocity is defined to be that constant velocity that gives the same displacement, for a given time period, as the actual, time varying velocity. $\endgroup$ – Alfred Centauri Jul 16 '13 at 21:55
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A simple way to see this is that the average velocity is the area under the velocity-time curve divided by the time interval. When you have constant acceleration, the area is a trapezium, and the result follows.

$$v_{\operatorname{avg}}=\frac{\mbox{Area of trapezium}}{\delta t}=\frac{\frac{1}{2}\left(v_0+v_f\right) \delta t}{\delta t}=\frac{v_0+v_f}{2}$$

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  • $\begingroup$ Don't see Dimension how that is a trapezium not triangle. $\endgroup$ – Bobby Jul 16 '13 at 20:57
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    $\begingroup$ Check the Wolfram|Alpha link. Because the function (velocity) has a non - zero value (the initial velocity) when time is 0. $\endgroup$ – Abhimanyu Pallavi Sudhir Jul 17 '13 at 4:36
  • $\begingroup$ This is a off topic question, but I can't find any other good online sources for that - what is heuristic ? $\endgroup$ – user77648 Apr 23 '16 at 2:42
  • $\begingroup$ Yup, almost (I know it intuitively to some degree, but I don't know the answer to some questions like why the are of rectangle is $height \; \times \; width$ or similar questions). But how does that relate to the notion of heuristic ? $\endgroup$ – user77648 Apr 23 '16 at 11:53

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