Question

An electron moving along the x-axis has a position given by $x = 20te^{-t}$ m, where $t$ is in second. How far is the electron from the origin when it momentarily stops?

• A. 20 m
• B. 20 em
• C. $\frac{20}{e}$ m
• D. zero

Hint:

For any function of form \( t^n e^{-at} \), the maximum value or stationary point always occurs at \( t = n/a \). Here \( n=1 \) and \( a=1 \), so \( t = 1 \) immediately without full differentiation steps.

Answer 1

The Correct Option is C

The problem requires us to find the position of an electron moving along the x-axis given by the equation \(x = 20te^{-t}\) when it momentarily stops. Here, \(t\) represents time in seconds.

Step-by-step Solution:

1. To determine when the electron momentarily stops, we need to find when its velocity is zero. Velocity \(v\) is the derivative of position \(x\) with respect to time \(t\).
2. Differentiate the given position function with respect to \(t\): \(\frac{dx}{dt} = \frac{d}{dt}(20te^{-t})\).
3. Use the product rule to find the derivative: \((uv)' = u'v + uv'\), where \(u = 20t\) and \(v = e^{-t}\).
4. Calculate the derivatives:
   • \(u' = \frac{d}{dt}(20t) = 20\)
   • \(v' = \frac{d}{dt}(e^{-t}) = -e^{-t}\)
5. Substitute the derivatives back into the product rule: \(\frac{dx}{dt} = 20 \cdot e^{-t} + 20t \cdot (-e^{-t})\).
6. Simplify the expression: \(\frac{dx}{dt} = 20e^{-t} - 20te^{-t} = 20e^{-t}(1-t)\).
7. Set the velocity to zero to find when the electron stops: \(20e^{-t}(1-t) = 0\).
8. As \(e^{-t}\) is never zero, set \(1 - t = 0\) which gives \(t = 1\).
9. Substitute \(t = 1\) back into the position equation to find the position at this time: \(x = 20 \times 1 \times e^{-1} = \frac{20}{e}\).

Therefore, the electron is \(\frac{20}{e}\) meters away from the origin when it momentarily stops. Thus, the correct option is:

Option 3: \(\frac{20}{e}\) m