uniform acceleration

Its velocity increases 4m/s every sec.

So during the first second, it starts with a velocity of zero, ends with a velocity of 4m/s. Average velocity during that second is 2m/s.

At the end of one second: distance traveled is 2 meters, ending velocity is 4m/2

check by equation:

d(t)=1/2at^2 +vt
d(1) =1/2(4m/s^2)*(1s)^2 +(0)(1s)
d(1)=2m

During the second second of motion, it starts with a velocity of 4m/2, ends with a velocity of 8m/s, for an average velocity of 6m/s. Distance during the second second is 6m, for a total distance of 8m.

Check by equation:

d(t)=1/2at^2 + vt

d(2) = 1/2 (4m/s^2) * (2s)^2+0(2s)
d=1/2(4)(4)m = 8m

At the end of 10 seconds:

d(10) = 1/2(4m/s^2)*(10s)^2 + 0(10s)
d(10) = 1/2(4m/s^2)*100s^2
d(100)=200m