College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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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 \( -406250 \, \text{m/s}^2 \).](https://content.bartleby.com/qna-images/question/a023f3c3-3186-4c61-aa74-30f5825e9d0f/f95e6c7a-e76c-4419-b4d2-c402f0509662/wdph7q.jpeg)
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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