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Chapter 2: Problem 104

Calc Air drag plays a significant role in the free fall of objects near Earth's surface. Suppose the acceleration of a falling object is given by the following function $$ a(v)=g-\alpha v \quad \text { (down is positive) } $$ where \(\alpha\) is a positive constant. (a) By integrating, find the velocity of a falling object as a function of time. (b) Find the terminal velocity of an object that falls from rest starting at \(t=0\).

Short Answer

Expert verified
The velocity of the falling object as a function of time is given by \(v(t) = (g/\alpha) * (1 - e^{-\alpha t})\). The terminal velocity is \(v_{terminal} = g/\alpha\).

Step by step solution

01

Integrate the acceleration function

We start by solving the integral of the given function \(a(v)\). Considering that \(a(v) = dv/dt\), we can rewrite the equation like this: \(dv/dt = g - \alpha v\). Separate variables and integrate. The integration gives us \(\int \frac{dv}{g -\alpha v} = \int dt\).
02

Solve the integral

Utilizing basic integral concepts, we get \(-1/\alpha * ln|g - \alpha v| = t + C\), where \(C\) is the constant of integration.
03

Find the constant of integration

We can use the initial condition to solve for the constant \(C\). When \(t = 0\), \(v = 0\) because the object starts from rest. So, substituting \(v = 0\) and \(t = 0\) in our equation, we get \(-1/\alpha * ln|g| = C\). So, the solution for the integral becomes: \(-1/\alpha * ln|g - \alpha v| = t - 1/\alpha * ln|g|\).
04

Express velocity as a function of time

By rearranging the terms and solving for \(v\), we get: \(v(t) = (g/\alpha) * (1 - e^{-\alpha t})\).
05

Find the terminal velocity

The terminal velocity \(v_{terminal}\) is achieved when the object stops accelerating, in other words when the time \(t\) approaches infinity. Hence, \(v_{terminal} = \lim_{t \to \infty} v(t)\). Calculating this limit we find: \(v_{terminal} = g/\alpha\).

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Key Concepts

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

Acceleration
In the context of a falling object, acceleration describes how quickly the velocity of that object changes over time as it descends towards the Earth. Our exercise specifies the acceleration function as \( a(v) = g - \alpha v \), where \( g \) represents the gravitational acceleration 鈥� typically around 9.8 m/s虏 near the Earth's surface.
  • The term \( \alpha v \) indicates the deceleration effect of air resistance, proportional to the velocity \( v \) and a constant \( \alpha \).
  • When a falling object moves fast enough, air resistance grows larger, thus decreasing net acceleration.
As velocity increases, the air resistance increases too, causing acceleration to decrease. When air resistance equals gravitational force, acceleration becomes zero. This happens at terminal velocity.
Terminal Velocity
Terminal velocity is a key concept in understanding the motion of objects through air. It is the constant speed an object ultimately reaches when the forces of gravity and air resistance balance out, resulting in zero net acceleration.
  • In our solution, when time \( t \) approaches infinity, velocity \( v(t) \) becomes a constant \( g/\alpha \), representing terminal velocity.
  • This value occurs because as speed builds up, drag forces due to air resistance oppose the gravitational pull more effectively.
  • Terminal velocity is crucial, as it determines the maximum speed a falling object can achieve in a medium like air.
For different objects, terminal velocity varies based on mass, shape, and air density, impacting real-world applications such as skydiving or the handling of small objects.
Integration
Integration is a fundamental mathematical process used to find quantities like velocity in this exercise. To find velocity as a function of time from the given acceleration, we need to integrate the function.
  • Starting with \( a(v) = dv/dt = g - \alpha v \), we separate variables to integrate, leading to: \( \int \frac{dv}{g - \alpha v} = \int dt \).
  • Integration allows us to reconstruct the velocity-time relationship from the acceleration equation.
  • After integrating, by applying our initial conditions (object starting from rest), we solve for the constant of integration, \( C \).
The solution to this integration gives us a clear formula for velocity over time: \( v(t) = (g/\alpha) \times (1 - e^{-\alpha t}) \). This key result helps us predict object velocity during free fall.
Free Fall
Free fall describes the motion of objects under the sole influence of gravity. In an ideal scenario with no air resistance, objects accelerate at \( g \) constantly. However, in practical scenarios near the Earth's surface, air resistance profoundly affects this motion.
  • We consider the effect of air on freely falling objects with the equation \( a(v) = g - \alpha v \).
  • This equation characterizes the net acceleration by factoring in gravity and resisting forces from air friction.
  • With increasing velocity, air resistance reduces the net acceleration until the object reaches terminal velocity.
Understanding free fall with air resistance in mind helps not only in academic exercises but also in designing technologies like parachutes, which control descent speed effectively.

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Most popular questions from this chapter

-Calc The velocity of a rocket is given by the following function: $$ v(t)=\left\\{\begin{array}{cc} 125\left(1-e^{-0.12 f}\right) & 022 s \end{array}\right\\} $$ (a) What are the SI units of the numerical values 125 , \(0.12\), and \(116.1\) ? (b) Calculate the speed of the rocket at \(t=2.2 \mathrm{~s}\) and at \(t=25 \mathrm{~s}\). (c) Calculate the displacement of the rocket from 0 to \(10 \mathrm{~s}\). (d) Calculate the displacement of the rocket from 0 to \(30 \mathrm{~s}\). SSM

\(\bullet \bullet\) A rocket is launched straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for \(10.0\) and \(5.0 \mathrm{~s}\), respectively, with no time interval between them. In Stage 1, the rocket fuel provides an upward acceleration of \(15 \mathrm{~m} / \mathrm{s}^{2}\). In Stage 2 , the acceleration is \(12 \mathrm{~m} / \mathrm{s}^{2}\). Neglecting air resistance, calculate the maximum altitude above the surface of Earth of the payload and the time required for it to return back to the surface. SSM

Biology A black mamba snake has a length of \(4.3\) \(\mathrm{m}\) and a top speed of \(8.9 \mathrm{~m} / \mathrm{s}\) ! Suppose a mongoose and a black mamba find themselves nose to nose. In an effort to escape, the snake accelerates at \(18 \mathrm{~m} / \mathrm{s}^{2}\) from rest. (a) How much time does it take for the snake to reach its top speed? (b) How far does the snake travel in that time? (c) Has the tail of the snake gone past the mongoose or does the mongoose have a chance to catch the black mamba?

Derive the equation that relates position to speed and acceleration but in which the time variable does not appear. Start with the basic equation for the definition of acceleration, \(a=\left(v-v_{0}\right) / t\), solve for \(t\), and substitute the resulting expression into the position versus time equation, \(x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}\).

The slope at a point on a position versus time graph of an object is A. the object's speed at that point. B. the object's instantaneous velocity at that point. C. the object's average velocity over a time interval centered on that point. D. the object's instantaneous acceleration at that point. E. the object's average acceleration over a time interval centered on that point.
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