Chapter 5 - Newton's Laws of Motion - Problems and Conceptual Exercises - Page 146: 79

(C) $1.05 \times 10^4 N$

We know that $a=\frac{v^2-v_{\circ}^2}{2s}$ We plug in the known values to obtain: $a=\frac{(0m/s)^2-(18m/s)^2}{2(1m)}$ $a=-162m/s$ Now $F=(65Kg)(-162m/s^2)$ $F=-1.05\times 10^4N$ Thus, the magnitude of force on the driver is $1.05\times 10^4N$ and hence the correct option is (C).