A bullet is fired through a board 12.1 cm thick, with a line of motion perpendicular to the face of the board. If the bullet enters with a speed of 375 m/s and emerges with a speed of 205 m/s, what is its acceleration as it passes through the board? (Take the direction the bullet is moving to be the positive.)
Vf^2=Vi^2+2ad solve for a
To find the acceleration of the bullet as it passes through the board, we can use the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
In this case, the bullet enters the board with an initial velocity of u = 375 m/s and emerges with a final velocity of v = 205 m/s. The displacement, s, is given as the thickness of the board, which is 12.1 cm.
First, let's convert the thickness of the board to meters:
s = 12.1 cm = 0.121 m
Now we can rearrange the kinematic equation to solve for acceleration:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (205^2 - 375^2) / (2 * 0.121)
Calculating the numerator:
205^2 - 375^2 = 42025 - 140625 = -98600
Substituting back into the equation:
a = (-98600) / (2 * 0.121)
Calculating the denominator:
2 * 0.121 = 0.242
Now we can find the acceleration:
a = -98600 / 0.242
a ≈ -406611.57 m/s²
Therefore, the acceleration of the bullet as it passes through the board is approximately -406611.57 m/s². The negative sign indicates that the acceleration is in the opposite direction of the bullet's motion.