A bullet is fired through a board, 14.0 cm thick, with its line of motion perpendicular to the face of the board. If it enters with a speed of 450 m/s and emerges with a speed of 220 m/s, what is the acceleration (in km per second squared) of the bullet as it passes through the board?

I know a = l/t^2, but there is no time given here.

Any reasoning behind the answer is GREATLY appreciated.

V(f)^2=V(i)^2+2*a*d; solve for a

.22(km/s)^2=.45(km/s)^2+2*a*.00014 km

.0484(km/s)^2=.2025(km/s)^2 + .00028(km)*a

-.1541(km/s)^2= .00028(km)*a

a= -550.36 km/(s^2)

thank you sir or madam.