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

Step 1: Understanding the Concept:
The velocity of a particle is the first derivative of its position with respect to time.
The term "momentarily stop" indicates that the instantaneous velocity of the electron becomes zero at a specific instant of time.

Step 2: Key Formula or Approach:
Calculate velocity $v$ by differentiating $x$ with respect to $t$:
\[ v = \frac{dx}{dt} \]
Set $v = 0$ to find the time $t$ of stopping, then substitute this $t$ back into the original $x$ equation.

Step 3: Detailed Explanation:
Given position function: \( x = 20te^{-t} \)
Differentiating using the product rule \(\frac{d}{dt}(uv) = u\frac{dv}{dt} + v\frac{du}{dt}\):
\[ v = \frac{d}{dt}(20te^{-t}) = 20 \left[ t \cdot \frac{d}{dt}(e^{-t}) + e^{-t} \cdot \frac{d}{dt}(t) \right] \]
\[ v = 20 [ t \cdot (-e^{-t}) + e^{-t} \cdot 1 ] \]
\[ v = 20e^{-t} (1 - t) \]
Setting \( v = 0 \) to find the time of stopping:
\[ 20e^{-t} (1 - t) = 0 \]
Since \( 20e^{-t} \) cannot be zero for finite time, we have \( 1 - t = 0 \), which gives \( t = 1 \text{ second} \).
Now, calculate the position at \( t = 1 \text{ s} \):
\[ x = 20(1)e^{-1} = \frac{20}{e} \text{ m} \]

Step 4: Final Answer:
The electron is at a distance of $\frac{20}{e}$ m from the origin.