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You are trying to improve your shooting skills by shooting at a can on top of a fence post. You miss the can, and the bullet, moving at \(200 . \mathrm{m} / \mathrm{s}\), is embedded \(1.5 \mathrm{~cm}\) into the post when it comes to a stop. If constant acceleration is assumed, how long does it take for the bullet to stop?

Short Answer

Expert verified
Answer: The bullet takes approximately 0.00015 seconds to come to a stop after hitting the fence post.

Step by step solution

01

List the given values and the required variable

Initial velocity (v₀): 200 m/s Final velocity (v): 0 m/s (since the bullet stops) Distance traveled (s): 0.015 m (1.5 cm converted to meters) Acceleration (a): unknown Time (t): required variable
02

Choose the correct equation of motion

Since we need to find the time taken to stop, we will use the equation of motion that relates initial velocity, final velocity, acceleration, and time: v = v₀ + at
03

Rearrange the equation to solve for time (t)

We can rearrange the equation in Step 2 to make time (t) the subject: t = (v - v₀) / a
04

Substitute values and solve for acceleration (a)

We need to find the acceleration first, so we will use another equation of motion that relates initial velocity, final velocity, acceleration, and the distance traveled: v² = v₀² + 2as Rearrange this equation to solve for acceleration (a): a = (v² - v₀²) / 2s Substitute the given values: a = (0² - (200)²) / (2 * 0.015) Calculate the acceleration: a ≈ -1,333,333.33 m/s² (since the bullet is decelerating)
05

Substitute values and solve for time (t)

Using the equation from Step 3, substitute the given values and the calculated acceleration to solve for time (t): t = (0 - 200) / (-1,333,333.33) Calculate the time taken for the bullet to stop: t ≈ 0.00015 s The bullet takes approximately 0.00015 seconds to come to a stop after hitting the fence post.

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Key Concepts

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

Constant Acceleration
Understanding constant acceleration is crucial when analyzing objects in motion. This concept implies that the rate of change in velocity is uniform over time. In simpler terms, the object gains or loses speed at a steady rate. For a bullet fired into a post, we assume constant acceleration to simplify the calculation.

When constant acceleration is assumed, the kinematic equations become powerful tools for solving motion problems. These equations can relate the initial and final velocities of the object, the time it takes for the object's velocity to change, the distance it travels during this time, and the constant acceleration itself. As seen in the textbook solution, these relationships are vital to determine the time it takes for the bullet to stop upon impact with the fence post.
Initial Velocity
The initial velocity (\( v_{0} \) ) is the speed at which an object starts its journey. It's an essential starting point for any motion analysis and is often given in problems as a known value. In our exercise, the bullet's initial velocity is 200 m/s, and this value is one of the key components used in the kinematic equations.

In practical terms, knowing the initial velocity allows us to predict how an object will behave under certain conditions of motion, especially when combined with other variables such as time and acceleration. For instance, it helps us understand how the bullet moves the moment it leaves the gun and before it begins to slow down upon impact.
Final Velocity
Final velocity (\( v \) ) is the speed at which an object ends its motion. For the bullet in our problem, the final velocity is 0 m/s, as it comes to a complete stop. Knowing the final velocity is crucial, especially when determining the overall change in an object's motion.

When calculating motion, the final velocity can be determined using the initial velocity, the acceleration, and the time taken to decelerate. The concept of final velocity is not just about the endpoint but about the entirety of the object's journey, reflecting how it was influenced by forces such as friction or, in this case, the resistance of the fence post material.
Kinematics
Kinematics is the branch of physics that describes the motion of points, objects, and groups of objects without considering the forces that caused the motion. It's concerned with variables like velocity, acceleration, time, and displacement.

The kinematic equations are the math expressions that link these variables together. They are a set of four equations which can be derived from the basic principles of motion. For our bullet example, kinematics tells us not only how long the bullet takes to stop, but also how its motion evolves throughout its flight. The use of these equations is evident in our solved problem, indicating the depth to which kinematics is integrated into understanding motion in physics.

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Most popular questions from this chapter

A jet touches down on a runway with a speed of \(142.4 \mathrm{mph}\). After \(12.4 \mathrm{~s},\) the jet comes to a complete stop. Assuming constant acceleration of the jet, how far down the runway from where it touched down does the jet stand?

Runner 1 is standing still on a straight running track. Runner 2 passes him, running with a constant speed of \(5.1 \mathrm{~m} / \mathrm{s}\). Just as runner 2 passes, runner 1 accelerates with a constant acceleration of \(0.89 \mathrm{~m} / \mathrm{s}^{2} .\) How far down the track does runner 1 catch up with runner \(2 ?\)

A runner of mass 56.1 kg starts from rest and accelerates with a constant acceleration of \(1.23 \mathrm{~m} / \mathrm{s}^{2}\) until she reaches a velocity of \(5.10 \mathrm{~m} / \mathrm{s}\). She then continues running at this constant velocity. How long does the runner take to travel \(173 \mathrm{~m} ?\)

Bill Jones has a bad night in his bowling league. Feeling disgusted when he gets home, he drops his bowling ball out the window of his apartment, from a height of \(63.17 \mathrm{~m}\) above the ground. John Smith sees the bowling ball pass by his window when it is \(40.95 \mathrm{~m}\) above the ground. How much time passes from the time when John Smith sees the bowling ball pass his window to when it hits the ground?

A train traveling at \(40.0 \mathrm{~m} / \mathrm{s}\) is headed straight toward another train, which is at rest on the same track. The moving train decelerates at \(6.0 \mathrm{~m} / \mathrm{s}^{2}\), and the stationary train is \(100.0 \mathrm{~m}\) away. How far from the stationary train will the moving train be when it comes to a stop?

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