/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 To study damage to aircraft that... [FREE SOLUTION] | 91Ó°ÊÓ

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To study damage to aircraft that collide with large birds, you design a test gun that will accelerate chicken-sized objects so that their displacement along the gun barrel is given by \(x=\) \(\left(9.0 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\right) t^{2}-\left(8.0 \times 10^{4} \mathrm{m} / \mathrm{s}^{3}\right) t^{3} .\) The object leaves the end of the barrel at \(t=0.025 \mathrm{s}\) (a) How long must the gun barrel be? \((b)\) What will be the speed of the objects as they leave the end of the barrel? (c) What net force must be exerted on a \(1.50-k g\) object at \((\mathrm{i}) t=0\) and (ii) \(t=0.025 \mathrm{s} ?\)

Short Answer

Expert verified
(a) 4.375 m, (b) 300 m/s, (c) (i) 27,000 N, (ii) 9,000 N.

Step by step solution

01

Calculate the displacement at t = 0.025 s

To find the length of the gun barrel, we need to determine the displacement of the object at the time it exits. Substitute \( t = 0.025 \) s into the displacement equation:\[ x = (9.0 \times 10^3) t^2 - (8.0 \times 10^4) t^3 \]\[ x = (9.0 \times 10^3) (0.025)^2 - (8.0 \times 10^4) (0.025)^3 \]Calculating, we get:\[ x = 9.0 \times 10^3 \times 0.000625 - 8.0 \times 10^4 \times 0.000015625 \]\[ x = 5.625 - 1.25 \]\[ x = 4.375 \text{ meters} \]Therefore, the gun barrel must be 4.375 meters long.
02

Determine the speed at t = 0.025 s

The speed of the object can be found by differentiating the displacement function with respect to time to get the velocity function. The velocity function is: \[ v(t) = \frac{d}{dt} [(9.0 \times 10^3) t^2 - (8.0 \times 10^4) t^3] \]\[ v(t) = 2 \times 9.0 \times 10^3 \times t - 3 \times 8.0 \times 10^4 \times t^2 \]\[ v(t) = 1.8 \times 10^4 \times t - 2.4 \times 10^5 \times t^2 \]Substitute \( t = 0.025 \) s:\[ v(0.025) = 1.8 \times 10^4 \times 0.025 - 2.4 \times 10^5 \times (0.025)^2 \]\[ v(0.025) = 450 - 150 \]\[ v(0.025) = 300 \text{ m/s} \]The speed of the objects as they leave the barrel is 300 m/s.
03

Calculate the net force at t = 0

The net force can be calculated using Newton's second law: \( F = ma \). First we find the acceleration by differentiating the velocity function.The acceleration function is: \[ a(t) = \frac{d}{dt} [1.8 \times 10^4 \times t - 2.4 \times 10^5 \times t^2] \]\[ a(t) = 1.8 \times 10^4 - 2 \times 2.4 \times 10^5 \times t \]\[ a(t) = 1.8 \times 10^4 - 4.8 \times 10^5 \times t \]At \( t = 0 \), \[ a(0) = 1.8 \times 10^4 \text{ m/s}^2 \]With \( m = 1.50 \text{ kg} \), the force is:\[ F(0) = 1.50 \times 1.8 \times 10^4 \]\[ F(0) = 2.7 \times 10^4 \text{ N} \]The net force exerted at \( t = 0 \) is 27,000 N.
04

Calculate the net force at t = 0.025 s

Again using \( F = ma \), substitute \( t = 0.025 \) s into the acceleration function:\[ a(0.025) = 1.8 \times 10^4 - 4.8 \times 10^5 \times 0.025 \]\[ a(0.025) = 1.8 \times 10^4 - 1.2 \times 10^4 \]\[ a(0.025) = 6.0 \times 10^3 \text{ m/s}^2 \]With \( m = 1.50 \text{ kg} \), the force is:\[ F(0.025) = 1.50 \times 6.0 \times 10^3 \]\[ F(0.025) = 9.0 \times 10^3 \text{ N} \]The net force exerted at \( t = 0.025 \) s is 9,000 N.

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Key Concepts

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

Projectile Motion
Projectile motion describes the movement of objects thrown or fired into the air, like a ball or, in this exercise, a chicken-sized object. These objects follow a curved path due to gravitational forces. The key to understanding projectile motion is to break it into two components: horizontal and vertical motion.

The horizontal motion remains constant because, under ideal conditions, there is no acceleration in this direction. In the vertical direction, the only force acting is gravity, leading to a constant acceleration of approximately 9.81 m/s² downward.

This problem involves finding the projectile's speed and distance at a specific time, which combines these concepts. Here, the gun accelerates the projectile until it leaves the barrel, forming part of the preliminary phase before true projectile motion takes place.
Newton's Laws of Motion
In this exercise, Newton's Laws of Motion play a central role in determining the forces acting on the projectile. Newton's Second Law of Motion states that force equals mass times acceleration, or \( F = ma \).

This law is crucial when calculating the net forces acting on the object at different instance times. Initially, a higher force is needed to overcome inertia and accelerate the object until it exits the barrel, reaching a high speed. As calculated in the solution, the force required varies with time depending on the acceleration observed at that specific moment.

Newton's second law connects directly to this change in forces, highlighting that a larger force is necessary at the initial stages of motion to achieve the desired velocity in a short time.
Calculus in Physics
Calculus is a powerful tool in physics for deriving equations of motion and resolving rates of change. This exercise involves using calculus to find velocity and acceleration from displacement.

The displacement function, \( x(t) \), given as a function of time, allows us to derive velocity by differentiating once with respect to time. The result is the velocity function which reveals the speed at any moment.

Further, differentiating the velocity function gives the acceleration, a key factor reflecting how fast the velocity of the object increases with time. With these derivative functions, not only can we find instantaneous velocities and accelerations, but they enable us to apply Newton's laws effectively to determine forces.
Kinematics
Kinematics deals with the description of motion without considering the forces causing it. It involves variables like displacement, velocity, and acceleration.

In kinematics, understanding equations that define motion over time is key. Here, the given displacement equation serves as a starting point to unravel a story of movement. By employing calculus, students can transition from this equation to derive velocity and acceleration functions, thus gaining insight into how motion progresses.

This method is fundamental in physics to predict motion paths and outcomes without external force considerations initially. By embracing the principles of kinematics, one can closely examine each stage of an object's journey, from rest to motion, in a very descriptive way.

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Most popular questions from this chapter

An electron (mass \(=9.11 \times 10^{-31} \mathrm{kg} )\) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 \(\mathrm{cm}\) away. It reaches the grid with a speed of \(3.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) . If the accelerating force is constant, compute (a) the acceleration; \((b)\) the time to reach the grid; (c) the net force, in newtons. (You can ignore the gravitational force on the electron.)

An object with mass \(m\) moves along the \(x\) -axis. Its position as a function of time is given by \(x(t)=A t-B t^{3},\) where \(A\) and \(B\) are constants. Calculate the net force on the object as a function of time.

An object of mass \(m\) is at rest in equilibrium at the origin. At \(t=0\) a new force \(\vec{F}(t)\) is applied that has components $$ F_{x}(t)=k_{1}+k_{2} y \quad F_{y}(t)=k_{3} t $$ where \(k_{1}, k_{2},\) and \(k_{3}\) are constants. Calculate the position \(\vec{r}(t)\) and velocity \(\vec{v}(t)\) vectors as functions of time.

A dockworker applies a constant horizontal force of 80.0 \(\mathrm{N}\) to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 \(\mathrm{m}\) in 5.00 \(\mathrm{s}\) . (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 \(\mathrm{s}\) s, how far does the block move in the next 5.00 \(\mathrm{s?}\)

A small \(8.00-\mathrm{kg}\) rocket burns fuel that exerts a time-varying upward force on the rocket. This force obeys the equation \(F=A+B t^{2} .\) Measurements show that at \(t=0,\) the force is \(100.0 \mathrm{N},\) and at the end of the first \(2.00 \mathrm{s},\) it is 150.0 \(\mathrm{N} .\) (a) Find the constants \(A\) and \(B,\) including their SI units. (b) Find the net force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after fuel ignition. (c) Suppose you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

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