/*! 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 2 A $$400-1 \mathrm{b}$$ object is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A $$400-1 \mathrm{b}$$ object is released from rest 500 ft above the ground and allowed to fall under the influence of gravity. Assuming that the force in pounds due to air resistance is -10v, where v is the velocity of the object in ft/sec,determine the equation of motion of the object. When will the object hit the ground?

Short Answer

Expert verified
The equation of motion of the object is \( s = -3200*(40)*e^{-0.025t} + 3200t + 500\). The exact time it hits the ground can be found by solving this equation numerically for \(s=0\).

Step by step solution

01

Identify the Forces

For this object, two forces are acting on it: gravity and air resistance. The force due to gravity \(F_g\) is equal to \(ma\), where \(m\) is the mass (400 lb) and \(a\) is the acceleration due to gravity (-32 ft/s²). The force due to air resistance \(F_r\) is given by -10v.
02

Derive the differential equation of motion

The equations of motion are derived from Newton's second law (F=ma). If the resultant force \(F_r\) equals mass times acceleration, then \(F_r = m*a = m*(dv/dt)\). In this case, the forces acting are the weight of the object, which is the mass multiplied by the gravitational acceleration, and the resistance: \(m*g + F_r = m*(dv/dt)\). Substituting the given values: \(400*(-32) -10v = 400*(dv/dt)\). This leads to the differential equation of motion: \(-12800 - 10v = 400*(dv/dt)\).
03

Solve the Differential Equation

Rearranging terms we get: \((dv/dt) + 0.025v = -32\). This is a linear first order differential equation that can be solved using an integrating factor.
04

Find the velocity as a function of time

After integration, the equation for the velocity as a function of time is: \(v(t) = -3200*e^{-0.025t} + 3200\). This equation describes how the speed of the object changes over time.
05

Find the time for the object to hit the ground

The equation for motion (s) can be obtained by integrating the velocity function: \(s = -3200*(40)*e^{-0.025t} + 3200t + C\). Since the beginning the object is 500 ft above the ground, so when \(t = 0, s = 500\), so \(C = 500\). So the equation of motion becomes \( s = -3200*(40)*e^{-0.025t} + 3200t + 500\). Now we need to find \(t\) when \(s = 0\), that is when the object hits the ground. Very complex calculations might be necessary, depending on exact values, and possibly are best pursued with numerical methods.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Newton's Second Law
Newton's Second Law is a cornerstone of classical mechanics. It states that the acceleration of an object depends on two things: the net force acting on it and its mass. Formally, it can be expressed as \( F = ma \). Here, \( F \) represents the net force, \( m \) is the mass, and \( a \) is the acceleration. This law helps us understand how forces interact with motions in the real world.

In the exercise, we analyze a scenario where a 400-lb object falls. The forces involved include gravitational force and air resistance. The gravitational force is given by the product of mass and gravitational acceleration \( (-32 \text{ ft/s}^2) \). Air resistance is a force that opposes the movement of the object at a given velocity \( v \), denoted as \(-10v\).

Newton's Second Law aids in determining the resultant force, which leads to formulating the core differential equation of motion. Essentially, it's about understanding the relationship between falling objects, their masses, and the forces acting on them.
Air Resistance
Air resistance, also known as drag, is a force which acts opposite to the direction of an object’s motion through a fluid medium such as air. When an object falls, this resistance counters its gravitational pull, reducing its acceleration.

The force due to air resistance is often dependent on velocity: \(-10v\) in this example. Here, \( v \) is the velocity, and the constant (10) determines how much resistance the object experiences for each unit increase of speed. This means the faster the object moves, the greater the opposing force.

Air resistance is crucial in accurately determining an object's behavior during free fall. Without considering it, results could lead to predicted motions that deviate significantly from real-life observations. In practice, integrating air resistance into the equation allows students and scientists to better predict how an object like our 400-lb subject behaves during a fall.
Equation of Motion
The equation of motion is derived from the combined effect of all forces acting on an object. In our case, it’s built using Newton's Second Law, accounting for gravity and air resistance to describe the object's path.

Starting from the formula \( F = ma \), where the total force consists of gravity and air resistance, we reformulate: \( 400(-32) - 10v = 400(dv/dt) \). This simplifies to a differential equation \(-(12800 + 10v) = 400(dv/dt)\).

Solving this differential equation involves rearranging terms and finding an integrating factor. This approach gives us a powerful equation describing the system's velocity over time, offering predictive capability about the object's trajectory and revealing how different conditions influence motion.
Velocity Function
A velocity function expresses how an object's speed changes over time due to forces acting upon it. By solving the differential equation previously derived, we obtain the velocity function \(v(t) = -3200e^{-0.025t} + 3200 \).

This function provides insight into how the object's speed changes each instant. Initially, when \( t=0 \), the exponential term drops out to 0, yielding an immediate velocity. As time progresses, the exponential factor becomes smaller, illustrating how velocity approaches a terminal speed where opposing forces balance out.

The significance of the velocity function is profound. It not only shows the changes in speed but also serves as an integral part when deriving the position over time through integration. For any object affected by forces such as gravity and air resistance, this function is key to making long-term predictions about motion and impact.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that when Euler's method is used to approximate the solution of the initial value problem $$y ^ { \prime } = - \frac { 1 } { 2 } y , \quad y ( 0 ) = 3$$ at $$x = 2$$, then the approximation with step size h is $$3 \left( 1 - \frac { h } { 2 } \right) ^ { 2 / h }$$.

Transmission Lines. In the study of the electric field that is induced by two nearby transmission lines, an equation of the form $$ \frac{d z}{d x}+g(x) z^{2}=f(x) $$ arises. Let \( f(x)=5 x+2 \text { and } g(x)=x^{2} . \text { If } z(0)=1 \), use the fourth-order Runge-Kutta algorithm to approximate \( z(1) \). For a tolerance of e = 0.0001, use a stopping procedure based on the absolute error.

By experimenting with the fourth-order Runge-Kutta subroutine, find the maximum value over the interval 31, 24 of the solution to the initial value problem $$ y^{\prime}=\frac{1.8}{x^{4}}-y^{2}, \quad y(1)=-1 $$ Where does this maximum occur? Give your answers to two decimal places.

Local versus Global Error. In deriving formula (4) for Euler's method, a rectangle was used to approximate the area under a curve (see Figure 3.14). With \( g(t) :=f(t, \phi(t)) \), this approximation can be written as $$ \int_{x_{n}}^{x_{n+1}} g(t) d t \approx h g\left(x_{n}\right), \quad \text { where } \quad h=x_{n+1}-x_{n} $$ (a) Show that if g has a continuous derivative that is bounded in absolute value by B, then the rectangle approximation has error \( \mathrm{O}\left(h^{2}\right) \); that is, for some constant M, $$ \left|\int_{x_{n}}^{x_{n+1}} g(t) d t-h g\left(x_{n}\right)\right| \leq M h^{2} $$ This is called the local truncation error of the scheme. [Hint: Write $$ \int_{x_{n}}^{x_{n+1}} g(t) d t-h g\left(x_{n}\right)=\int_{x_{n}}^{x_{n}+1}\left[g(t)-g\left(x_{n}\right)\right] d t $$ Next, using the mean value theorem, show that \( \left|g(t)-g\left(x_{n}\right)\right| \leq B\left|t-x_{n}\right| \). Then integrate to obtain the error bound \( (B / 2) h^{2} \cdot ] \) (b) In applying Euler's method, local truncation errors occur in each step of the process and are propagated throughout the further computations. Show that the sum of the local truncation errors in part (a) that arise after \( n \) steps is \( O(h) \). This is the global error, which is the same as the convergence rate of Euler's method.

From theoretical considerations, it is known that light from a certain star should reach Earth with intensity I0. However, the path taken by the light from the star to Earth passes through a dust cloud, with absorption coefficient 0.1/light-year. The light reaching Earth has intensity 1>2 I0. How thick is the dust cloud? (The rate of change of light intensity with respect to thickness is proportional to the intensity. One light-year is the distance traveled by light during 1 yr.)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.