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An indestructible bullet \(2.00 \mathrm{~cm}\) long is fired straight through a board that is \(10.0 \mathrm{~cm}\) thick. The bullet strikes the board with a speed of \(420 \mathrm{~m} / \mathrm{s}\) and emerges with a speed of \(280 \mathrm{~m} / \mathrm{s}\). (a) What is the average acceleration of the bullet through the board? (b) What is the total time that the bullet is in contact with the board? (c) What thickness of board (calculated to \(0.1 \mathrm{~cm}\) ) would it take to stop the bullet, assuming the acceleration through all boards is the same?

Short Answer

Expert verified
The bullet's average acceleration while inside the board is \( -7.00 \times 10^5 \, m/s^2 \), it was in contact with the board for \( 2.00 \times 10^{-4} \, s \), and it would require a board that is \( 63.0 \, cm \) thick to stop the bullet.

Step by step solution

01

Calculate the average acceleration

The change in velocity of the bullet \( \Delta v \) is the final velocity \( v_f \) minus the initial velocity \( v_i \), or \( \Delta v = v_f - v_i \). The average acceleration \( a \) is then the change in velocity divided by the time it takes for this change to occur, or \( a = \Delta v / \Delta t \). Since we know that the distance the bullet travels through the board \( d \) is the length of the board plus the length of the bullet \( d = d_b + d_l \), we can write a kinematic equation in terms of the total length of the board and the bullet \( d \), the initial velocity \( v_i \), the final velocity \( v_f \) and the acceleration \( a \). The equation \( v_f^2 = v_i^2 + 2ad \) then allows us to solve for \( a \).
02

Calculate the time of contact

Now that we have the acceleration, we can use it to find out how long the bullet was in contact with the board. The kinematic equation \( v_f = v_i + at \) allows us to to solve for \( t \). Here, \( t \) is the time the bullet was in contact with the board.
03

Calculate the board thickness needed to stop the bullet

Finally, we find out how thick the board would need to be to stop the bullet. We set the final velocity \( v_f \) to zero and the recalculate the distance \( d_b \) required using the kinematic equation \( v_f^2 = v_i^2 + 2ad \)

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Key Concepts

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

Kinematic Equations
Kinematic equations are the backbone of classic mechanics, describing the motion of objects based on their initial velocity, acceleration, and time elapsed. They provide a mathematical framework to analyze motion in one dimension. One such equation is \( v_f^2 = v_i^2 + 2ad \), where \( v_f \) stands for the final velocity, \( v_i \) is the initial velocity, \( a \) represents acceleration, and \( d \) is the distance traveled.

In the context of our exercise, we used this kinematic equation to determine the average acceleration (\( a \)) of a bullet as it passed through a board. By understanding the kinematic equations, we can predict and calculate various aspects of an object's journey, whether it’s the distance a bullet travels after penetrating a board, or the height a ball reaches before it comes to a stop mid-air.
Average Acceleration
Average acceleration is a measure of how quickly an object changes its velocity. It's the rate at which velocity increases or decreases over time. Given by the formula \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) represents the change in velocity and \( \Delta t \) denotes the time span for this change.

When we apply this to the bullet problem, we infer that the bullet’s velocity decreased from \( 420 \, m/s \) to \( 280\, m/s \) while passing through the board. To find the average acceleration, we would normally need the time – which could be found using another kinematic equation. This concept is vital as it can also indicate the force that was exerted on the bullet by the board, assuming we know the bullet's mass.
Velocity-Time Relationship
The velocity-time relationship is a key concept in physics, involving how velocity changes over time under constant acceleration. Expressed mathematically by the equation \( v_f = v_i + at \), it allows us to calculate one of these variables when the other three are known. In this relationship, \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is acceleration, and \( t \) is time.

The solution to our bullet problem uses this relationship. Given the bullet's initial speed, final speed, and average acceleration, we can determine the time the bullet was in contact with the board. This becomes crucial when assessing the impact duration of a moving object, such as in car crash tests or sports physics.
Bullet Penetration Physics
Bullet penetration physics examines how a projectile interacts with a target material, considering factors like projectile velocity, material resistance, and energy transfer. In our exercise, we discuss this in terms of how the bullet's velocity changes as it passes through a board and how thickness affects the bullet's ability to penetrate.

The kinematic equations model this interaction by letting us solve for the board thickness required to stop the bullet. Assuming constant acceleration, we can infer the distance at which the bullet will come to rest—providing valuable insights for bulletproof materials development, forensic ballistics, and safety standards in materials testing. This real-world application underscores the importance of the theoretical framework provided by kinematic equations in solving practical problems.

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Most popular questions from this chapter

. A tortoise can run with a speed of \(0.10 \mathrm{~m} / \mathrm{s}\), and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for \(2.0\) minutes. The tortoise wins by a shell \((20 \mathrm{~cm}) .\) (a) How long does the race take? (b) What is the length of the race?

A person takes a uip, driving with a constant speed of \(89.5 \mathrm{~km} / \mathrm{h}\), except for a \(22.0\) -min rest stop. If the person's average speed is \(77.8 \mathrm{~km} / \mathrm{h}\), how much time is spent on the trip and how far does the person travel?

A certain freely falling object requires \(1.50 \mathrm{~s}\) to travel the last \(80.0 \mathrm{~m}\) before it hits the ground. From what height above the ground did it fall?

A car is traveling due east at \(25.0 \mathrm{~m} / \mathrm{s}\) at some instant. (a) If its constant acceleration is \(0.750 \mathrm{~m} / \mathrm{s}^{2}\) due east, find its velocity after \(8.50 \mathrm{~s}\) have elapsed. (b) If its constant acceleration is \(0.750 \mathrm{~m} / \mathrm{s}^{2}\) due west, find its velocity after \(8.50 \mathrm{~s}\) have elapsed.

Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than \(800 \mathrm{~m} / \mathrm{s}^{2}\) lasting for any length of time will not cause injury, whereas an acceleration greater than \(1000 \mathrm{~m} / \mathrm{s}^{2}\) lasting for at least \(1 \mathrm{~ms}\) will cause injury. Suppose a small child rolls off a bed that is \(0.40 \mathrm{~m}\) above the floor. If the floor is hardwood, the child's head is brought to rest in approximately \(2.0 \mathrm{~mm}\). If the floor is carpeted, this stopping distance is increased Io about \(1.0 \mathrm{~cm}\). Calculate the magnitude and duration of the deceleration in both eases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.

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