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Chapter 3: Problem 51

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of \(6.20 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}\) for \(8.10 \times 10^{-4} \mathrm{s}\). What is its muzzle velocity (that is, its final velocity)?

Short Answer

Expert verified
The muzzle velocity of the bullet is approximately \(502 m/s\).

Step by step solution

01

Identify the givens

We are given: - Acceleration (a) = \(6.20 \times 10^5 m/s^2\) - Time (t) = \(8.10 \times 10^{-4} s\) - Initial velocity (u) = 0 (since the bullet is at rest)
02

Apply the equation of motion

We will use the following equation of motion to find the muzzle velocity: Final Velocity (v) = Initial Velocity (u) + Acceleration (a) 脳 Time (t) Substitute the given values into the equation: v = 0 + (\(6.20 \times 10^5 m/s^2\)) 脳 (\(8.10 \times 10^{-4} s\))
03

Calculate the muzzle velocity

Now, we can calculate the muzzle velocity: v = (\(6.20 \times 10^5 m/s^2\)) 脳 (\(8.10 \times 10^{-4} s\)) v 鈮� \(502 m/s\) The muzzle velocity of the bullet is approximately 502 m/s.

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Key Concepts

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

Understanding Acceleration
Acceleration is the rate at which an object changes its velocity. It is a vector quantity, which means it has both magnitude and direction. In the context of kinematics, acceleration is crucial because it helps us understand how quickly an object is speeding up or slowing down. For the case of the bullet in the gun, the average acceleration is given as \(6.20 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}\).
  • A positive acceleration indicates an increase in velocity.
  • A negative acceleration (also known as deceleration) indicates a decrease in velocity.
  • Units for acceleration are meters per second squared \(\mathrm{m}/\mathrm{s}^{2}\).
By knowing the acceleration, we can predict the change in velocity over time, which is essential for calculating the final velocity of the bullet as it travels through the barrel of the gun.
Understanding Equations of Motion
Equations of motion are mathematical formulas that help us calculate various kinematic quantities such as displacement, velocity, acceleration, and time. When an object moves with constant acceleration, as with the bullet in the gun, we can use these equations to determine different aspects of its motion. The equation used in the problem is: \[v = u + at\]where:
  • \(v\) is the final velocity, which we want to find.
  • \(u\) is the initial velocity. For a bullet starting from rest, \(u = 0\).
  • \(a\) is the acceleration.
  • \(t\) is the time period for which the acceleration is applied.
These equations allow us to link the known values to find unknown aspects of the motion. For instance, by substituting the given values of acceleration and time, one can calculate the final velocity of the bullet.
Velocity Calculation Made Simple
Velocity represents the speed of an object in a specific direction. Calculating it when an object starts from rest and is uniformly accelerated involves a straightforward process. In the example of the bullet, we need to calculate the final velocity, also known as muzzle velocity, which tells how fast the bullet is moving when it leaves the gun.Given:
  • Initial velocity \(u = 0\)
  • Acceleration \(a = 6.20 \times 10^5 \mathrm{m}/\mathrm{s}^2\)
  • Time \(t = 8.10 \times 10^{-4} \mathrm{s}\)
Plug these values into the equation:\[v = u + at\]Substitution yields:\[v = 0 + (6.20 \times 10^5) \times (8.10 \times 10^{-4})\]Calculating this gives us:\[v \approx 502 \mathrm{m}/\mathrm{s}\]This means the bullet's muzzle velocity is approximately 502 meters per second, indicating it leaves the barrel at a very high speed.

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Most popular questions from this chapter

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of \(11.0 \mathrm{m} / \mathrm{s}\). How long a time does he have to get out of the way if the shot was released at a height of \(2.20 \mathrm{m}\) and he is \(1.80 \mathrm{m}\) tall?

In a 100 -m race, the winner is timed at 11.2 s. The second-place finisher's time is 11.6 s. How far is the second-place finisher behind the winner when she crosses the finish line? Assume the velocity of each runner is constant throughout the race.

Blood is accelerated from rest to \(30.0 \mathrm{cm} / \mathrm{s}\) in a distance of \(1.80 \mathrm{cm}\) by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat?

A cyclist sprints at the end of a race to clinch a victory. She has an initial velocity of \(11.5 \mathrm{m} / \mathrm{s}\) and accelerates at a rate of \(0.500 \mathrm{m} / \mathrm{s}^{2}\) for \(7.00 \mathrm{s}\). (a) What is her final velocity? (b) The cyclist continues at this velocity to the finish line. If she is 300 m from the finish line when she starts to accelerate, how much time did she save? (c) The second-place winner was \(5.00 \mathrm{m}\) ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at \(11.8 \mathrm{m} / \mathrm{s}\) until the finish line. What was the difference in finish time in seconds between the winner and runner-up? How far back was the runner-up when the winner crossed the finish line?

A particle moves in a straight line with an initial velocity of \(30 \mathrm{m} / \mathrm{s}\) and constant acceleration \(30 \mathrm{m} / \mathrm{s}^{2}\). (a) What is its displacement at \(t=5 \mathrm{s} ?\) (b) What is its velocity at this same time?
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