Answer:
[tex]-4.40\cdot 10^{5} m/s^2[/tex]
Explanation:
We can assume that the bullet decelerates only when it is passing through the board, from the moment it starts touching the board until it leaves it completely; so, it accelerates for a total distance of
d = 2.27 cm + 8 cm = 10.27 cm = 0.103 m
So we can use the following equation to find its acceleration:
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 265 m/s is the final speed of the bullet
u = 401 m/s is its initial speed
a is the acceleration
d = 0.103 m
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{265^2-401^2}{2(0.103)}=-4.40\cdot 10^{5} m/s^2[/tex]
