Question

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

Answer 1

This is a kinematics problem.
Let's list out the kinematic formulas:
$x= v_{i} t+at^2/2$
$v_{f} ^2= v_{i} ^2+2ax$
$v_{f} = v_{i} +at$

What do we know?
We have vi, vf, and x. We're trying to find acceleration. The formula that has all four of these is the second. Now, let's plug in and solve for a.
$v_{f} ^2= v_{i} ^2+2ax$
$220^2=450^2+2*.14*a$
$-154100=.28*a$
$a=-5.50*10^5 m/s^2$