Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 51: 55

$3.8\times10^{4}~m/s^2$

The required acceleration can be determined as follows: $a=|\frac{v_f^2-v_{\circ}^2}{2\Delta x}|$ We plug in the known values to obtain: $a=|\frac{(0m/s)^2-(130m/s)^2}{2(0.22m)}|$ $a=3.8\times 10^4m/s^2$