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An electron moving along the \(x\) axis has a position given by \(x=16 t e^{-t} \mathrm{~m}\), where \(t\) is in seconds. How far is the electron from the origin when it momentarily stops?

Short Answer

Expert verified
The electron is approximately 5.88 meters from the origin when it stops.

Step by step solution

01

Identify When the Electron Stops

To find when the electron momentarily stops, we need to find when its velocity is zero. The velocity is the derivative of the position function with respect to time: \ \(v(t) = \frac{d}{dt}[16t e^{-t}]\).
02

Differentiate the Position Function

Apply the product rule to differentiate the function. If \(u = 16t\) and \(v = e^{-t}\), then \ \(\frac{d}{dt}[16t e^{-t}] = u'v + uv'\). \ Here, \(u' = 16\) and \(v' = -e^{-t}\), so \ \(v(t) = 16e^{-t} + 16t(-e^{-t}) = 16e^{-t} - 16te^{-t}\).
03

Set the Velocity to Zero

Set the expression for the velocity to zero to find the time when the electron stops: \ \(16e^{-t} - 16te^{-t} = 0\). Factor out \(16e^{-t}\): \ \(16e^{-t}(1-t) = 0\).
04

Solve for Time

The equation \(16e^{-t}(1-t) = 0\) results in two possible solutions: \(e^{-t} = 0\) or \(1 - t = 0\). Since \(e^{-t} > 0\) for all \(t\), it follows \(1 - t = 0\), giving \(t = 1\).
05

Substitute Time into Position Function

Substitute \(t = 1\) into the position function to find the electron's distance from the origin: \ \(x(1) = 16 \times 1 \times e^{-1} = \frac{16}{e}\, \mathrm{m}.\)
06

Calculate the Distance

Compute the distance using the value \(e \approx 2.718\): \ \(x(1) \approx \frac{16}{2.718} \approx 5.88 \, \mathrm{m}.\)

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Key Concepts

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

Electron Motion
In the world of physics, understanding the motion of electrons is vital for grasping the basics of kinematics. Kinematics is the branch of mechanics that focuses on the motion of objects without considering the forces that cause such motion.
In this scenario, an electron is moving along the x-axis, and its position at any given time is described by the equation \(x = 16t e^{-t} \). This tells us where the electron is located at any point in time \(t\).
Here are some insights regarding electron motion:
  • The position equation incorporates both linear and exponential components, highlighting an interesting relationship between time and the electron's position.
  • Observing the behavior of this function helps us predict when the electron stops or changes direction.
  • In practice, knowing the exact position or motion of electrons is crucial in fields like electronics and atomic physics.
By studying the motion of electrons, we unlock a greater understanding of electric currents, semiconductor behavior, and the fundamental processes in nature at a microscopic level.
Velocity Calculation
The concept of velocity is central to understanding how the electron's motion changes over time. Velocity is simply the rate of change of position with respect to time. To find the electron's velocity, we need to differentiate the position function.
Using our position function \(x = 16t e^{-t}\), we calculate the velocity \(v(t)\) by differentiating the position:
  • Apply the product rule: if \(u = 16t\) and \(v = e^{-t}\), then the derivative \(\frac{d}{dt}[uv] = u'v + uv'\).
  • Calculate \(u' = 16\) and \(v' = -e^{-t}\).
  • This gives us the velocity \(v(t) = 16 e^{-t} - 16t e^{-t}\).
Understanding the calculated velocity equation helps us identify moments in time when the electron is not moving. At these points, the velocity equals zero, informing us about the electron's momentary stops or transitions in motion.
Position Function Differentiation
When solving problems related to motion, differentiation is a crucial mathematical tool. Differentiating a position function reveals the velocity at each instant, thus providing key insights into the dynamics of the motion. In this exercise, the position of the electron is given by \(x(t) = 16t e^{-t}\), and differentiating this function unveils how its position changes over time.
To differentiate:
  • Recognize it as a product of two functions: \(16t\) and \(e^{-t}\).
  • Use the product rule for differentiation \(\frac{d}{dt}[u \, v]= u'v + uv'\).
  • Here, \(u = 16t\), \(u' = 16\), \(v = e^{-t}\), and \(v' = -e^{-t}\).
  • Thus, the derivative is \(16 e^{-t} - 16t e^{-t}\).
The result of this differentiation gives us the velocity function, which we set to zero to find the exact time \(t\) when the electron stops. Mastering differentiation of position functions is essential in physics, as it allows us to understand velocity and acceleration, further enhancing our ability to predict and analyze motion accurately.

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Most popular questions from this chapter

An object falls a distance \(h\) from rest. If it travels \(0.60 h\) in the last \(1.00 \mathrm{~s}\), find (a) the time and (b) the height of its fall. (c) Explain the physically unacceptable solution of the quadratic equation in \(t\) that you obtain.

You are arguing over a cell phone while trailing an unmarked police car by \(25 \mathrm{~m}\); both your car and the police car are traveling at \(120 \mathrm{~km} / \mathrm{hr}\). Your argument diverts your attention from the police car for \(2.0 \mathrm{~s}\) (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that \(2.0 \mathrm{~s}\), the police officer begins braking suddenly at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another \(0.40\) s to realize your danger and begin braking. (b) If you too brake at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\), what is your speed when you hit the police car?

The position function \(x(t)\) of a particle moving along an \(x\) axis is \(x=4.0-6.0 t^{2}\), with \(x\) in meters and \(t\) in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph \(x\) versus \(t\) for the range \(-5 \mathrm{~s}\) to \(+5 \mathrm{~s}\). (f) To shift the curve rightward on the graph, should we include the term \(+20 t\) or the term \(-20 t\) in \(x(t) ?(g)\) Does that inclusion increase or decrease the value of \(x\) at which the particle momentarily stops?

The brakes on your car can slow you at a rate of \(5.2 \mathrm{~m} / \mathrm{s}^{2}\). (a) If you are going \(146 \mathrm{~km} / \mathrm{h}\) and suddenly see a state trooper, what is the minimum time in which you can get your car under the \(90 \mathrm{~km} / \mathrm{h}\) speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.) (b) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for such a slowing.

A stone is dropped into a river from a bridge \(53.6 \mathrm{~m}\) above the water. Another stone is thrown vertically down \(1.00 \mathrm{~s}\) after the first is dropped. The stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released.

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