Distance covered by a brick in $1$ second time interval 
A brick is dropped from the roof of a tall building.After it has been
  falling for a few seconds ,it falls $40.0$ meters in a $1.00$-s time
  interval.What distance will it fall during the next $1.00$ seconds?
  Ignore air resistance.

I've seen on yahoo answer that the solution provided for second part of the question,namely the distance covered in the second interval of $1.00$ second,can be found just by realizing that the midpoint of this interval occurs $1$ second later the midpoint of the first interval(so this would allow us to calculate instantaneous velocity).
However I am a bit skeptical ,because after the midpoint of the first interval the brick's velocity kept raising in magnitude,so I would think that  it hasn't necessarily have to be at the midpoint of the second interval exactly $1$ second later after the first one.
This way I would be assuming that the brick's velocity is constant over that interval,while it isn't.
Another point of confusion is that I can find many intervals of $1$ second where the brick falled a distance of $40$ meters..
Question
Can you guys make this clear ?Am I right or wrong ?If I am right how would I solve the problem ?

More simply, the SUVAT equation $v = u + at$ shows that the "final" velocity increases by $a\; (= g)$ every second, thus so must the average velocity.(**)
Hence distance travelled in the next second  $ = (40 +g)\cdot1 = 40+g$
(**) Explanation :
Velocities at time, $0,1,2$ seconds: $u, u+g, u+2g$
Average velocity $0-1: u + 0.5g = 40 =$ distance travelled in first second
Average velocity $1-2: u + 1.5g =40 +g =$ distance travelled in next second 
