1. boomerang285

2. drewinator3

At 8 seconds the problem has been reduced to a linear equation. $X_1 = X + Vt + {1 \over 2} a t^2$ where $X = 1.9m, V = 1.7 {m \over s}, a = 0 {m \over s^2}$ Then this equation becomes... $X(t) = 1.9 + 1.7t$ Before the 8 second mark none of them seem to be linear or parabolic. This is just speculation though. I believe this because it seems that the velocity and acceleration are irregular functions of time, the functions wouldn't fit a linear curve or the perfect definition of a parabolic conic section. With out the explicit functions I couldn't know for sure whether this is the case though

3. Mashy

yea its very difficult to make out what curve that is.. but surely neither linear nor parabolic.. but you can approximate it to be quite linear you know.. you see acceleration is almost zero..