Distance formula and addition of meters per s

[booga]

Hello guys.

I have issues understanding a rather simple physics concept and I hope you can help me out. I've tried asking help from several people, but nobody was able to isolate the problem.
For constant acceleration, it is said that d=(Vi+Vf)/2*t. d being distance, Vi is initial velocity, Vf is final velocity and t is time of the travel.

The problem occurs when I try to add - what seems intuitively for me atleast - the distance of object traveled each second with the added velocity from the constant acceleration.

An example, object starts at 10m/s and after 10 seconds reach a velocity of 20m/s.
150 = (10+20)/2*10, and (20-10)/10 = 1, so acceleration is 1m/s/s.

Now when I try to add the distance traveled of each second over a duration of 10 seconds, I get 10+11+12+13+14+15+16+17+18+19 = 145. I don't add 20 because to my understanding the object doesn't actually travel any seconds with 20m/s, it merely reaches that velocity after going a full second at 19m/s.

So I can't the results to match each other, only if I do something like (10+19)/2*10.

 

[Doc Al]

Now when I try to add the distance traveled of each second over a duration of 10 seconds, I get 10+11+12+13+14+15+16+17+18+19 = 145.

Use the same formula for each 1-second interval that you used for the 10-second interval:
In the first second, d = (10 + 11)/2 * 1 = 10.5 (not 10). And so on, for each interval.

 

[booga]

I see. So just to pinpoint where my intuition is wrong; that by adding the way I did, I assumed the acceleration only occurs after each second. It's more of a constant "push" toward the object, that happens during the seconds itself.

Thanks for the help.