(A) Kinetic energy is calculated as K = mv^{2}/2 where m is the mass of the object and v is its velocity. The initial kinetic energy is then K_{0} = mv_{0}^{2}/2 = (12100 kg)(15 m/s)^{2}/2 = 1361250 J. Since the car came to a stop, its final velocity is v = 0 and thus it has no kinetic energy. The final kinetic energy is K = 0 J.

(B) The change in kinetic energy, ΔK, is simply ΔK = K - K_{0} = 0 J - 1361250 J = -1361250 J.

(C) The change in kinetic energy is equal to the amount of work done. Thus, we take our result from (B) such that W = ΔK = -1361250 J.

(D) We have that the work is equal to the product of the force applied and the displacement over which the force is applied (W = FΔd). Solving the equation for the displacement, we see that Δd = W/F. Taking the work from (C) and the given applied force of F = 18150 N, we see that the car slowed to a stop over a distance |Δd| = |W/F| = |(-1361250 J)/(18150 N)| = |-75 m| = 75 m.