A bullet is fired through a board, 0.28 cm thick, with its line of motion perpendicular to the face of the board. If it enters with a speed of 50 m/s and emerges with a speed of 15 m/s, what is the bullet's acceleration in m/s as it

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**Bullet Through a Board Physics Problem**

*A bullet is fired through a board, 0.28 cm thick, with its line of motion perpendicular to the face of the board. If it enters with a speed of 50 m/s and emerges with a speed of 15 m/s, what is the bullet's acceleration in m/s² as it passes through the board?*

To solve this problem, we need to apply the principles of kinematics. We are given the following information:

- Initial speed of the bullet, \( v_i = 50 \) m/s
- Final speed of the bullet, \( v_f = 15 \) m/s
- Thickness of the board, \( d = 0.28 \) cm \( = 0.0028 \) m

Using the kinematic equation:
\[ v_f^2 = v_i^2 + 2ad \]

Where:
- \( v_f \) is the final velocity
- \( v_i \) is the initial velocity
- \( a \) is the acceleration
- \( d \) is the distance traveled through the board

We need to solve for the acceleration \( a \).

Rearranging the formula to solve for \( a \):
\[ a = \frac{v_f^2 - v_i^2}{2d} \]

Plugging in the given values:
\[ a = \frac{(15)^2 - (50)^2}{2 \cdot 0.0028} \]
\[ a = \frac{225 - 2500}{0.0056} \]
\[ a = \frac{-2275}{0.0056} \]
\[ a \approx -406250 \, \text{m/s}^2 \]

The negative sign indicates that the bullet is decelerating as it passes through the board. The bullet's acceleration is approximately \( -406250 \, \text{m/s}^2 \).
Transcribed Image Text:**Bullet Through a Board Physics Problem** *A bullet is fired through a board, 0.28 cm thick, with its line of motion perpendicular to the face of the board. If it enters with a speed of 50 m/s and emerges with a speed of 15 m/s, what is the bullet's acceleration in m/s² as it passes through the board?* To solve this problem, we need to apply the principles of kinematics. We are given the following information: - Initial speed of the bullet, \( v_i = 50 \) m/s - Final speed of the bullet, \( v_f = 15 \) m/s - Thickness of the board, \( d = 0.28 \) cm \( = 0.0028 \) m Using the kinematic equation: \[ v_f^2 = v_i^2 + 2ad \] Where: - \( v_f \) is the final velocity - \( v_i \) is the initial velocity - \( a \) is the acceleration - \( d \) is the distance traveled through the board We need to solve for the acceleration \( a \). Rearranging the formula to solve for \( a \): \[ a = \frac{v_f^2 - v_i^2}{2d} \] Plugging in the given values: \[ a = \frac{(15)^2 - (50)^2}{2 \cdot 0.0028} \] \[ a = \frac{225 - 2500}{0.0056} \] \[ a = \frac{-2275}{0.0056} \] \[ a \approx -406250 \, \text{m/s}^2 \] The negative sign indicates that the bullet is decelerating as it passes through the board. The bullet's acceleration is approximately \( -406250 \, \text{m/s}^2 \).
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