I'm not sure if it has more acceleration or not, but even if it had the same acceleration (I assume we're talking about average acceleration over the entire movement, to preclude the idea of the acceleration ceasing earlier), and was traveling a longer distance, then (assuming the same static start), it would have to have higher velocity, because it would be under acceleration longer. If it has more acceleration and a longer distance, it has to have a higher velocity at the end.
That is not entirely correct. Consider this:
Acceleration/Formula
= average acceleration
=
final velocity
= starting velocity
= elapsed time
From the web
Acceleration (a) is the change in
velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast
velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
So, δV is always going to be faster than v-sub, that is just physics.
But if a mass is always accelerating (which is not possible) the velocity is null, at any point on a line. This is the reason for the v-sub variable. Yea, it confuses me too but it is a constant.
The easier way for me to think about it is to say an object can neither instantly start moving at a velocity nor can it instantly stop. This is part of being the vector quantity mentioned.
In motion control we often have to calculate final velocity from a 'flying' start. When considering the final destination (if relevant) the vector quantity usually has a higher value than δt since mass is complex and time is constant.
I think what you are saying is defined as ramp. The angle of acceleration. If you look at them on a plane you can much easier understand the difference between v and v-sub.
Hope this makes sense.
***Note: I can never get certain ASCII characters to work on this site.