Why Is Instantaneous Velocity Equal to Average Velocity?

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SUMMARY

The instantaneous velocity at the midpoint of a time interval is equal to the average velocity over that interval when acceleration is constant. This relationship arises from the definition of average velocity, which is calculated as the mean of the initial and final velocities, v_1 and v_2. The mathematical proof involves demonstrating that the time at which the average velocity is reached coincides with the midpoint of the time interval, leading to the conclusion that the average velocity is represented as (v_1 + v_2) / 2.

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Does anyone happen to know why the instantaneous velocity at the midpoint of a time interval is equal to the average velocity over the same time interval?? I can't seem to prove this reasoning.

Thanks! :smile:
 
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You're talking about when acceleration is constant... It's just from the definition of "average" - the 2 in the denominator is where you get the halfway point in time.
 
First note that the average velocity is midway between the two velocities values, [itex]v_1,\ v_2[/itex], at the endpoints of the interval. Since
[tex]v_1+\frac{1}{2}(v_2-v_1)=v_1+\frac{v_2}{2}-\frac{v_1}{2}[/tex]

[tex]=\frac{v_1+v_2}{2}[/tex]
We therefore need to show that the time at which the average velocity is reached is in the middle of the time interval. In the drawing time is on the horizontal x-axis and speed on the vertical y-axis. What needs to be proved then in the drawing is that [itex]AD=BC[/itex], It is clear that both these length are given by
[tex]\frac{\Delta v}{2\tan(\theta)}[/tex]
 

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