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Chapter 2: Problem 65

A bullet going \(310 \mathrm{~m} / \mathrm{s}\) strikes a 5.0 -cm-thick target. (a) What constant acceleration is required if the bullet is to stop within the target? (b) What's its acceleration if the bullet emerges from the target at \(50 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) -961000 m/s虏 (b) -936000 m/s虏.

Step by step solution

01

Identify Given Information

For part (a), the bullet's initial velocity \( v_i = 310 \text{ m/s} \), the final velocity \( v_f = 0 \text{ m/s} \), and the displacement \( s = 0.05 \text{ m} \). For part (b), the initial velocity remains \( v_i = 310 \text{ m/s} \), the final velocity after passing through the target is \( v_f = 50 \text{ m/s} \), and the displacement remains \( s = 0.05 \text{ m} \).
02

Use the Equation of Motion for Part (a)

We use the equation \( v_f^2 = v_i^2 + 2as \) for part (a). Substitute the known values: \( 0 = 310^2 + 2a(0.05) \).
03

Solve for Acceleration in Part (a)

Rearrange the equation to solve for \( a \): \( 0 = 96100 + 0.1a \). Hence, \( a = -961000 \text{ m/s}^2 \).
04

Use the Equation of Motion for Part (b)

For part (b), apply the same formula: \( v_f^2 = v_i^2 + 2as \). Substitute the known values: \( 50^2 = 310^2 + 2a(0.05) \).
05

Solve for Acceleration in Part (b)

Simplify the equation: \( 2500 = 96100 + 0.1a \). Solve for \( a \) by arranging: \( a = \frac{2500 - 96100}{0.1} = -936000 \text{ m/s}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the forces causing the motion. It focuses strictly on the geometry of motion. To analyze the motion of a bullet hitting a target, we utilize key variables such as:
  • Velocity: the speed of an object in a particular direction.
  • Acceleration: the rate of change of velocity.
  • Displacement: the change in position of an object.

In kinematic problems, knowing at least three out of these four variables allows us to determine the fourth, using kinematic equations. These are essential for solving motion-related problems, such as determining how a moving bullet interacts with a target.
Acceleration
Acceleration is a crucial concept in understanding kinematics. It represents how quickly an object's velocity changes. In the bullet problem, constant acceleration is assumed to simplify the solution. Here's why it matters:
  • A positive acceleration indicates an increase in velocity.
  • A negative acceleration, as in our bullet problem, indicates a decrease in velocity 鈥� often called deceleration or braking.

In the exercise, the bullet decelerates rapidly upon impacting the target, requiring us to calculate this deceleration using the kinematics formula. Calculating acceleration helps us understand how swiftly an object slows down, stops, or even changes direction.
Equations of Motion
Equations of motion are a set of formulas that allow us to calculate unknown variables of moving objects. They are derived from the relationships between displacement, velocity, acceleration, and time. One commonly used equation is:
\[ v_f^2 = v_i^2 + 2as \]
Where:
  • \( v_f \) is the final velocity
  • \( v_i \) is the initial velocity
  • \( a \) is the acceleration
  • \( s \) is the displacement
This formula is particularly useful when time is not a given factor. In our problem, we use this equation twice: first to find the acceleration required to stop the bullet, and secondly to find the acceleration when the bullet exits with a residual velocity. Mastering these equations allows us to dissect motion problems with ease and predict future motion based on initial conditions.

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Most popular questions from this chapter

A rocket accelerates straight up from the ground at \(12.6 \mathrm{~m} / \mathrm{s}^{2}\) for \(11.0 \mathrm{~s}\). Then the engine cuts off and the rocket enters free fall. (a) Find its velocity at the end of its upward acceleration. (b) What maximum height does it reach? (c) With what velocity does it crash to Earth? (d) What's the total time from launch to crash?

You're driving at \(12.8 \mathrm{~m} / \mathrm{s},\) and are \(16.0 \mathrm{~m}\) from an intersection when you see a stoplight turn yellow. What acceleration do you need in order to just stop before the intersection? (a) \(-0.8 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(-5.1 \mathrm{~m} / \mathrm{s}^{2} ;\) (c) \(-7.4 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(-10.2 \mathrm{~m} / \mathrm{s}^{2}\)

Brain in juries in auto accidents. Brain injuries generally occur any time the brain's acceleration reaches \(100 g\) for even a short time. Consider a car running into a solid barrier. With an airbag, the driver's head moves through a distance of \(20 \mathrm{~cm}\) while the airbag stops it. Without an airbag, the head continues forward until the seatbelt stops the torso, causing the head to stop in a distance of only \(5.0 \mathrm{~cm} .\) For each case, find the maximum speed with which the car can strike the barrier without causing brain injury.

A car is speeding at \(75 \mathrm{mi} / \mathrm{h}(33.4 \mathrm{~m} / \mathrm{s})\). A police cruiser starts in pursuit from rest when the car is \(100 \mathrm{~m}\) past the cruiser. At what rate must the cruiser accelerate to catch the speeder before the state line, \(1.2 \mathrm{~km}\) away from the speeding car?

Can an object have zero velocity yet nonzero acceleration? Give an example.
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