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Chapter 8: Problem 4

Find the distance which an object moves in time \(t\) if it starts from rest and has an acceleration \(d^{2} x / d t^{2}=g e^{-k t} .\) Show that for small \(t\) the result is approximately (1.10), and for very large \(t\), the speed \(d x / d t\) is approximately constant. The constant is called the terminal speed. (This problem corresponds roughly to the motion of a parachutist.)

Short Answer

Expert verified
The distance is \(x(t) = -\frac{g}{k^2} e^{-k t} + \frac{g}{k} t\). For small \(t\): \(x(t) \approx \frac{g}{2} t^2\). For large \(t\): \(v_{\text{term}} = \frac{g}{k}\).

Step by step solution

01

- Understand the Problem

We need to find the distance an object moves over time when starting from rest with a given acceleration. The acceleration is defined as \(\frac{d^{2} x}{d t^{2}} = g e^{-k t}\). Also, show the approximate distance for small and large times.
02

- Integrate the Acceleration to Get the Velocity

Integrate the acceleration with respect to time to find the velocity. \(\frac{d v}{d t} = g e^{-k t}\) gives us: \[v(t) = \int g e^{-k t} d t + C_1 \]. Since the object starts from rest, \(v(0) = 0\), we have: \[v(t) = -\frac{g}{k} e^{-k t} + \frac{g}{k}\]
03

- Integrate the Velocity to Get the Position

Integrate the expression for velocity to find the position \(x(t)\): \[x(t) = \int \left(-\frac{g}{k} e^{-k t} + \frac{g}{k}\right) d t + C_2 \]. Applying the initial condition \(x(0) = 0\), we get: \[x(t) = -\frac{g}{k^2} e^{-k t} + \frac{g}{k} t\]
04

- Analyze for Small Time (t Approaches 0)

For small \(t\), \(e^{-k t} \approx 1 - k t\). Thus, the position can be approximated by: \[x(t) \approx -\frac{g}{k^2} (1 - k t) + \frac{g}{k} t = \frac{g}{2} t^2\]
05

- Analyze for Large Time (t Approaches Infinity)

For very large \(t\), \(e^{-k t} \rightarrow 0\). Hence, the velocity approaches a constant value called terminal speed \(v_{\text{term}}\): \[v_{\text{term}} = \frac{g}{k}\]

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Key Concepts

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

Kinematics
  • Kinematics is the branch of physics that studies motion without considering the forces that cause it.
  • It focuses on quantities like displacement, velocity, and acceleration.
  • The fundamental equations of kinematics are used to describe the motion of objects.
The exercise involves calculating how far an object moves over time with a continuously changing acceleration. Since the object starts from rest, its initial position and velocity are zero. This ensures a straightforward integration process, connecting acceleration to velocity and then to position. Understanding how these quantities change with time is key to solving kinematic problems efficiently.
Acceleration
Acceleration describes how quickly an object's velocity changes. In this problem, the acceleration is given by \(\frac{d^{2} x}{d t^{2}} = g e^{-k t}\). Note that this acceleration is not constant; it decreases over time. To solve for distance, we first needed the velocity by integrating the acceleration with respect to time:
\[ v(t) = \frac{d x}{d t} = -\frac{g}{k} e^{-k t} + \frac{g}{k} \]
Starting from rest means initial velocity, \(v(0) = 0\), making our integration straightforward. Acceleration thus affects how the object speeds up or slows down, dictating its journey from starting point to a certain distance over given time intervals.
Integration
Integration is a fundamental mathematical process in physics. It allows us to find quantities like velocity and position from acceleration. In this exercise, we integrated the acceleration function to find the velocity and then integrated the velocity to determine the position.
First, we found velocity: \[\frac{d x}{d t} = v(t) = -\frac{g}{k} e^{-k t} + \frac{g}{k} \]
Next, integrating velocity to find the position:
\[ x(t) = \frac{g}{k^2}(1 - e^{-k t}) + C_2 \]
By applying initial conditions (starting from rest), we solved for any constants and simplified our expressions. Understanding and applying integration in these steps is crucial for translating how acceleration impacts velocity and, subsequently, position over time.

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Most popular questions from this chapter

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$(2 x+y) d y-(x-2 y) d x=0$$

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x\left(y y^{\prime \prime}+y^{\prime 2}\right)=y y^{\prime} \quad \text { Hint: Let } u=y y^{\prime}$$.

(a) Consider a light beam traveling downward into the ocean. As the beam progresses, it is partially absorbed and its intensity decreases. The rate at which the intensity is decreasing with depth at any point is proportional to the intensity at that depth. The proportionality constant \(\mu\) is called the linear absorption coefficient. Show that if the intensity at the surface is \(I_{0},\) the intensity at a distance \(s\) below the surface is \(I=I_{0} e^{-\mu s} .\) The linear absorption coefficient for water is of the order of \(10^{-2} \mathrm{ft}^{-1}\) (the exact value depending on the wavelength of the light and the impurities in the water). For this value of \(\mu,\) find the intensity as a fraction of the surface intensity at a depth of \(1 \mathrm{ft}\), 50 ft, 500 ft, 1 mile. When the intensity of a light beam has been reduced to half its surface intensity \(\left(I=\frac{1}{2} I_{0}\right),\) the distance the light has penetrated into the absorbing substance is called the half-value thickness of the substance. Find the half-value thickness in terms of \(\mu .\) Find the half-value thickness for water for the value of \(\mu\) given above. (b) Note that the differential equation and its solution in this problem are mathematically the same as those in Example 1, although the physical problem and the terminology are different. In discussing radioactive decay, we call \(\lambda\) the decay constant, and we define the half-life \(T\) of a radioactive substance as the time when \(N=\frac{1}{2} N_{0}\) (compare half-value thickness). Find the relation between \(\lambda\) and \(T.\)

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}-2 y^{\prime}+5 y=5 x+4 e^{x}(1+\sin 2 x)$$

Use the given solutions of the homogeneous equation to find a particular solution of the given equation. You can do this either by the Green function formulas in the text or by the method of variation of parameters in Problem \(14 \mathrm{b}\) $$x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=x \ln x ; \quad x, x^{2}$$
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