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
v is defined as the total distance
d traveled in some time
t1:
[MATH]\overline{v}=\frac{d}{t_1}[/MATH]
But, we also have (from the area under the linear velocity function over the interval
[0,t1], 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.
