A bullet is fired horizontally at 312 m/sec and when it enters a wooden block, it comes to a rest after travel

Question

Please explain:

A bullet is fired horizontally at 312 m/sec and when it enters a wooden block, it comes to a rest after traveling 4 cm in the wood. What is the magnitude and direction (relative to its initial velocity) of the bullet’s acceleration?

Answer 1

Final velocity squared - Initial velocity squared = 2 * acceleration * displacment
V^2 - U^2 = 2as

V = 0; U = 312 m/s; s = 0.04 m; a= ?

a = -U^2/(2s) = -(312 m/s) ^2/(2*(0.04 m)
a = -1.22*10^6 m/s^2