Q14P - Fundamentals Of Physics

Question

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

Step-by-Step Solution

Answer

The distance traveled by the electron is  5.9 m.

Step 1: Given data

The equation that governs the motion of the electron is given as,

 $\mathrm{x}={16}^{-\mathrm{t}}\mathrm{m}$

Step 2: Understanding the concept of velocity and displacement.

The velocity can be found by taking the derivative of x, to get the time at which electrons stop. Using this time in the equation for displacement, the distance traveled by an electron can be calculated.

Formula:

$V=\frac{dx}{dt}$

Step 3: Calculate the distance

Differentiating the given equation with respect to time, you can find

$$\begin{gathered} V=\frac{dx}{dt} \\ =\frac{d}{dt}\left(16t\right)\times e^{-t}+16t\times \frac{d}{dt}\left(e^{-t}\right) \end{gathered}$$

You get derivative,

$\mathrm{V}=16(1-\mathrm{t}){\mathrm{e}}^{-\mathrm{t}}$

You get V = 0, when t = 1 s, it means an electron stops at .

Plug this time, t = 1  in equation of position you get,

 $$\begin{gathered} \mathrm{x}=16\left(1\right){\mathrm{e}}^{-1} \\ =5.9 \mathrm{m} \end{gathered}$$

Therefore, an electron stops at distance 5.9 m.