**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 \( = 0.0028 \) mUsing 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 boardWe 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 \( -406250 \, \text{m/s}^2 \).

Given : thickness of board, t = 0.28 cm = 0.0028 m speed with which it enters, u = 50…

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

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

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Distance and Speed

The speed, distance and time relation forms the basis of study of motion. You will need these concepts in the study of Mechanics. This concept is applicable for students who are studying the following courses –

Question

**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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