Chapter 5 - Newton's Laws of Motion - Problems and Conceptual Exercises - Page 140: 3

$ 2.57\ m$

We know that $a=\frac{F}{m}$ $\implies a=\frac{10.1N}{12.3Kg}$ $a=0.8211m/s^2$ Now $\Delta x=v_{\circ}+\frac{1}{2}a\Delta t^2$ We plug in the known values to obtain: $\Delta x=\frac{1}{2}(0.8211m/s^2)(2.5s)^2$ $\Delta x=2.57m$