Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 51: 52

$2.83\times10^{5}~m/s^2$

$1~yd=3~ft$, $1~m=3.281~ft$ $v_0=0$ $v=(12,000yd/s)(\frac{3~ft}{1~yd})(\frac{1~m}{3.281~ft})=10,972~m/s$ $\Delta x=(700~ft)(\frac{1~m}{3.281~ft})=213~m$ $v^2=v_0^2+2a\Delta x$ $(10,972~m/s)^2=0^2+2a(213~m)$ $a=\frac{(10,972~m/s)^2}{2(213~m)}=2.83\times10^{5}~m/s^2$