Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 51: 49

(a) $x=0.90~m$ (b) $x=3.6~m$ (c) $x=8.1~m$

$x=x_0+v_0t+\frac{1}{2}at^2$. Now, assuming that the start position is at the origin ($x_0=0$): (a) $x=0+0\times(1.0~s)+\frac{1}{2}(1.8~m/s^2)(1.0~s)^2=0.9~m$ (b) $x=0+0\times(2.0~s)+\frac{1}{2}(1.8~m/s^2)(2.0~s)^2=3.6~m$ (c) $x=0+0\times(3.0~s)+\frac{1}{2}(1.8~m/s^2)(3.0~s)^2=8.1~m$