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Chapter 2: Problem 54

A bullet is fired through a board \(10.0 \mathrm{~cm}\) thick, with a line of motion perpendicular to the face of the board. If the bullet enters with a speed of \(400 . \mathrm{m} / \mathrm{s}\) and emerges with a speed of \(200 . \mathrm{m} / \mathrm{s}\), what is its acceleration as it passes through the board?

Short Answer

Expert verified
Answer: The acceleration of the bullet as it passes through the board is -300,000 m/s².

Step by step solution

01

Identify the given variables

We are given the following information: - Initial velocity of the bullet, \(v_i = 400 \,\text{m/s}\) - Final velocity of the bullet, \(v_f = 200 \,\text{m/s}\) - Thickness or displacement of the board, \(d = 10.0 \,\text{cm} = 0.1 \,\text{m}\) (converted to meters)
02

Determine the appropriate kinematic equation

Since we are given initial and final velocities, displacement, and we need to find the acceleration, we should use the following kinematic equation: \(v_f^2 = v_i^2 + 2ad\) We can rearrange this equation to solve for the acceleration, \(a\).
03

Rearrange the equation to solve for acceleration

\(a = \frac{v_f^2 - v_i^2}{2d}\)
04

Plug in the given values into the equation

\(a = \frac{(200 \,\text{m/s})^2 - (400 \,\text{m/s})^2}{2(0.1 \,\text{m})}\)
05

Calculate the acceleration

\(a = \frac{40000 \,\text{m}^2/\text{s}^2 - 160000 \,\text{m}^2/\text{s}^2}{0.2 \,\text{m}}\) \(a = -60000 \,\text{m}^2/\text{s}^2 \times \frac{1}{0.2 \,\text{m}}\) \(a = -300000 \,\text{m/s}^2\) The acceleration of the bullet as it passes through the board is \(-300,000 \,\text{m/s}^2\). The negative sign indicates that the bullet is decelerating as it goes through the board.

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