Oh, i think I got it. So when an object moves with constant velocity it means that Δυ=0 and in same time periods it moves same distances, meaning that in every moment it has the same instantaneous velocity, which is also same to the average. Right?
I guess that it doesn't only have to do with the acceleration, right? I mean that what I understand, is that the average velocity will always be between the highest and the lowest value of instantaneous velocity as the velocity is a continuous function, I can't explain it better so if I say something wrong please tell me.But I can't think of it in practice. Could you please give me an example of that?
Suppose the acceleration \(a\) is constant, so that the velocity \(v\) at time \(t\) is given by:
[MATH]v(t)=at+v_0[/MATH]
Now, the average velocity \(\overline{v}\) is defined as the total distance \(d\) traveled in some time \(t_1\):
[MATH]\overline{v}=\frac{d}{t_1}[/MATH]
But, we also have (from the area under the linear velocity function over the interval \([0,t_1]\), which forms a trapezoid):
[MATH]d=\frac{t_1}{2}(at_1+2v_0)[/MATH]
Hence:
[MATH]\overline{v}=\frac{\dfrac{t_1}{2}(at_1+2v_0)}{t_1}=\frac{1}{2}at_1+v_0[/MATH]
Equating this to the velocity function, there results:
[MATH]\frac{1}{2}at_1+v_0=at+v_0[/MATH]
[MATH]t=\frac{1}{2}t_1[/MATH]
And so as out intuition might expect, with constant acceleration, the instantaneous velocity is equal to the average velocity half way through the time interval.
