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Chapter 2: Problem 112

The speed of a bullet is measured to be \(640 \mathrm{~m} / \mathrm{s}\) as the bullct cmerges from a barrel of length \(1.20 \mathrm{~m}\). A ssuming constant accelcration, find the time that the bullet spends in the barrel after it is fired.

Short Answer

Expert verified
The bullet spends approximately 0.00375 seconds in the barrel.

Step by step solution

01

Identify the Known Variables

We have the initial velocity of the bullet as it starts moving in the barrel, which is \(u = 0 \ \text{m/s}\), because it starts from rest. The final velocity \(v\) is given as \(640 \ \text{m/s}\). The length of the barrel, which is the distance \(s\), is \(1.20 \ \text{m}\). We need to find the time \(t\) that the bullet spends in the barrel.
02

Use the Kinematic Equation

The kinematic equation we use here is \(v^2 = u^2 + 2as\), where \(a\) is the acceleration. Since \(u = 0\), the equation simplifies to \(v^2 = 2as\). Plugging in the known values, we have \((640)^2 = 2a \times 1.20\).
03

Solve for Acceleration \(a\)

Rearrange the equation from Step 2 to solve for \(a\): \[ a = \frac{(640)^2}{2 \times 1.20} \]Calculate \(a\) to get \(a = 170666.67 \ \text{m/s}^2\).
04

Use the Time Equation

Now, use the kinematic equation \(v = u + at\) to solve for \(t\). Since \(u = 0\), it simplifies to \(v = at\), so \(t = \frac{v}{a}\).
05

Solve for Time \(t\)

Substitute the value of \(v\) and \(a\) into the equation from Step 4:\[ t = \frac{640}{170666.67} \]Calculate \(t\) to find that the time the bullet spends in the barrel is approximately \(0.00375 \ \text{seconds}\).

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Key Concepts

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

Constant Acceleration
Constant acceleration refers to a situation where the rate of change of velocity is uniform during the entire motion. This is a crucial concept in understanding many physical phenomena, especially in kinematics. In this context, when a bullet moves through a barrel and emerges at a known speed of 640 m/s, we assume it experiences a constant force over the barrel's length.
Understanding constant acceleration helps simplify our analysis of motion. It allows us to use specific kinematic equations that relate acceleration, initial velocity, final velocity, distance, and time.
There are some key points to keep in mind:
  • Kinematic equations are only valid under constant acceleration conditions.
  • By knowing the final speed and the acceleration, we can backtrack to find other unknowns, such as time.
  • In our scenario, since the bullet starts from rest, its initial velocity is 0 m/s.
Bullet Speed
The speed of a bullet as it exits the barrel is a fascinating measurement that can provide insight into the gun's power and design. In our exercise, the bullet reaches a high speed of 640 m/s over a short distance of 1.20 m. This speed is determined by its acceleration throughout the barrel.
Bullet speed matters because it influences:
  • Accuracy and precision: Faster speeds often yield straighter, more predictable trajectories.
  • Impact force: Greater speed translates to more kinetic energy upon impact.
From a kinematic standpoint, bullet speed connects closely with acceleration. By understanding how speed grows uniformly (under constant acceleration), we can solve related problems easily, like finding the time spent in motion.
Motion Analysis
Motion analysis is the process of examining the movement of objects using fundamental physics principles. In this exercise, we analyze the bullet's motion using kinematic equations, which provide a framework for calculating vital variables like time, distance, speed, and acceleration.
Steps to conduct motion analysis:
  • Identify known variables: For instance, initial velocity, final velocity, and distance.
  • Choose suitable equations: Kinematic equations are applied based on the known and unknown variables.
  • Solve for the unknowns: These might include finding time or acceleration as the bullet travels.
Through this structured analysis, we determine that the bullet's time inside the barrel is approximately 0.00375 seconds. By applying motion analysis techniques, complicated problems become more manageable, allowing for accurate predictions and solutions.

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Most popular questions from this chapter

A stone is dropped into a river from a bridge \(43.9 \mathrm{~m}\) above the water. Another stone is thrown vertically down \(1.00 \mathrm{~s}\) after the first is dropped. The stones strike the water at the same timc. (a) What is the initial speed of the scoond stonc? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released.

In \(1889,\) at Jubbulpore, India, a tug-of-war was finally won after \(2 \mathrm{~h} 41 \mathrm{~min},\) with the winning team displacing the center of the rope \(3.7 \mathrm{~m} .\) In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?

Two particles move along an \(x\) axis. The position of particle 1 is given by \(x=6.00 t^{2}+3.00 t+2.00\) (in meters and seconds); the acceleration of particle 2 is given by \(a=-8.00 t\) (in meters per second squared and seconds) and, at \(t=0,\) its velocity is \(20 \mathrm{~m} / \mathrm{s}\). When the velocities of the particles match, what is their velocity?

A motorcyclist who is moving along an \(x\) axis directed toward the east has an acceleration given by \(a=(6.1-1.2 t) \mathrm{m} / \mathrm{s}^{2}\) for \(0 \leq t \leq 6.0 \mathrm{~s}\). At \(t=0,\) the velocity and position of the cyclist are \(2.7 \mathrm{~m} / \mathrm{s}\) and \(7.3 \mathrm{~m}\). (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel between \(t=0\) and \(6.0 \mathrm{~s} ?\)

A car traveling \(56.0 \mathrm{~km} / \mathrm{h}\) is \(24.0 \mathrm{~m}\) from a barrier when the driver slams on the brakes. The car hits the barrier \(2.00 \mathrm{~s}\) later. (a) What is the magnitude of the car's constant acceleration before impact? (b) How fast is the car traveling at impact?
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