Question

A body of mass \(100 \mathrm{~g}\) is dropped from rest toward the earth from a height of \(1000 \mathrm{~m}\). As it falls, air resistance acts upon it, and this resistance (in newtons) is proportional to the velocity \(v\) (in meters per second). Suppose the limiting velocity is \(245 \mathrm{~m} / \mathrm{sec}\). (a) Find the velocity and distance fallen at time \(t\) secs. (b) Find the time at which the velocity is one-fifth of the limiting velocity.

Short answer

The velocity function is \(v(t) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t})\) and the distance fallen function is \(s(t) = -\frac{mg}{k^2}(kt + me^{-\frac{k}{m}t} - m)\). Based on the limiting velocity, we find that \(\frac{k}{m} = \frac{g}{245}\). The time at which the velocity is one-fifth of the limiting velocity can be calculated by solving \(v(t) = (1/5)(245) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t})\) for t.

Step-by-step solution

Determine the forces acting on the body

Based on the given information, there are two forces acting on the body: the force due to gravity and the force due to air resistance, which is proportional to its velocity. Let m be the mass of the body, g be the acceleration due to gravity, and k be the proportionality constant representing air resistance. We can write these forces as: Force due to gravity = \(mg\) Force due to air resistance = \(kv\) The net force acting on the body will be the difference between these two forces.

Apply Newton's second law of motion

Newton's second law of motion states that the net force acting upon an object is equal to the mass of the object multiplied by its acceleration (F = ma). We can write the equation for the net force acting on the body as: \(ma = mg - kv\) Let's solve for the acceleration: \(a = g - \frac{k}{m}v\)

Set up the differential equation relating velocity and acceleration

Acceleration is the derivative of velocity with respect to time. Thus, we can write the equation as: \(\frac{dv}{dt} = g - \frac{k}{m}v\)

Solve the differential equation for velocity

Let's solve the first-order linear differential equation for velocity: First, find the integrating factor, which is: \(e^{\int -\frac{k}{m} dt}\) The constant k/m can be found using the limiting velocity information. When the velocity reaches its limiting value, the acceleration becomes zero. Therefore, we have: \(245 = g - \frac{k}{m}(245)\) Now, solve for k/m. Next, multiply both sides of the differential equation by the integrating factor and integrate to find the velocity function: \(v(t) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t})\)

Calculate distance fallen using the velocity function

To find the distance fallen (s), we need to integrate the velocity function with respect to time: \(s(t) = \int v(t) dt\) Integrating the velocity function, we have: \(s(t) = -\frac{mg}{k^2}(kt + me^{-\frac{k}{m}t} - m)\)

Calculate the time for one-fifth of the limiting velocity

To find the time when the velocity is one-fifth of the limiting velocity, we set the velocity function equal to one-fifth times the limiting velocity, and solve for t: \(v(t) = (1/5)(245) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t})\) Solve for t.

Key concepts

Newton's Laws of Motion

Newton's Laws of Motion are fundamental principles in physics. They describe how objects move and interact under the influence of various forces. In this problem, Newton's second law of motion is particularly useful. This law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration: \( F = ma \).

In the context of the exercise, a body is falling under the influence of gravity and air resistance. The force of gravity acting on the body is calculated using the formula \( mg \), where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity (approximately \( 9.8 \ m/s^2 \)).

The other force at play is air resistance, which acts in the direction opposite to the motion of the body. It is proportional to the body's velocity and is represented as \( kv \), where \( k \) is the proportionality constant. Therefore, the net force is the force of gravity minus the force of air resistance: \( mg - kv \).

Using Newton's second law, we can relate this net force to its acceleration. This sets up the equation \( ma = mg - kv \) to determine the body's movement.

Air Resistance Modeling

Air resistance, also known as drag, is a force that opposes the motion of an object through air. This resistance increases with the speed of the object. In this exercise, air resistance is modeled as proportional to the velocity of the body.

When an object moves through air, it experiences a resistive force that is proportional to its velocity, often referred to as viscous drag. The expression for air resistance is given by \( kv \), where \( k \) is a constant factor that represents the extent of the air resistance experienced by the body. This resistance force works against the force of gravity, slowing the object down.

Modeling air resistance is crucial for predicting the object's motion because it affects how quickly the object reaches a steady-state fall, known as the limiting velocity. The differential equation that describes the velocity of the object includes this resistive term, making the problem a more realistic depiction of falling motion.

Limiting Velocity

Limiting velocity, also known as terminal velocity, is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration. In this scenario, it's when air resistance equals the force of gravity.

The concept arises when the net force on the object becomes zero. This means the gravitational force \( mg \) is perfectly balanced by the air resistance \( kv \), resulting in no further acceleration, and the object continues to fall at this constant speed. Mathematically, this is described as \( mg = kv_{lim} \), where \( v_{lim} \) is the limiting velocity.

In this exercise, we know the limiting velocity is \( 245 \ m/s \). To find the proportionality constant \( k \), you can set up the equation \( mg = kv_{lim} \) and solve for \( k \). Limiting velocity helps us understand and predict the behavior of objects in motion where air resistance is a factor, crucial for calculating the velocity function accurately.

Linear Differential Equation

Linear differential equations are a type of equation involving derivatives, which describe how a quantity changes over time. They are called "linear" because they depend linearly on the function and its derivatives.

In this problem, we deal with a first-order linear differential equation to describe the falling body's velocity with respect to time: \( \frac{dv}{dt} = g - \frac{k}{m}v \). This represents the relationship between the velocity's rate of change (acceleration) and the forces acting on the body.

To solve such equations, we often use methods like finding an integrating factor. It's a systematic approach to simplifying the equation into a form that can be integrated. The process involves identifying this integrating factor and applying it to the equation to find the velocity function \( v(t) \).

Finally, the velocity function \( v(t) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t}) \) provides a mathematical expression that describes how the velocity progresses over time. Linear differential equations are pivotal in modeling real-world scenarios where rates of change are influenced by multiple variables.