An electron along the x - axis has a position given by `x = 20te^(-t)m` where t is in  second.  The distance of the electron from the origin when it momentarily stops, is

Options

A. 120m
B. 20m
C. `(20)/(e )m`
D. zero

Explanation

To solve the problem, we need to find the distance of the electron from the origin when it momentarily stops. The position of the electron is given by the equation:

\[ x(t) = 20t e^{-t} \]

### Step-by-Step Solution:

**Step 1: Find the velocity of the electron.**

The velocity \( v(t) \) is the derivative of the position function \( x(t) \) with respect to time \( t \). We can use the product rule for differentiation since \( x(t) \) is a product of two functions: \( 20t \) and \( e^{-t} \).

Using the product rule:

\[ v(t) = \frac{dx}{dt} = \frac{d(20t)}{dt} \cdot e^{-t} + 20t \cdot \frac{d(e^{-t})}{dt} \]

Calculating the derivatives:

\[ \frac{d(20t)}{dt} = 20 \]
\[ \frac{d(e^{-t})}{dt} = -e^{-t} \]

Now substituting back into the equation:

\[ v(t) = 20 e^{-t} + 20t (-e^{-t}) = 20 e^{-t} (1 - t) \]

**Step 2: Set the velocity to zero to find when the electron stops.**

To find the time when the electron momentarily stops, we set the velocity \( v(t) \) to zero:

\[ 20 e^{-t} (1 - t) = 0 \]

Since \( e^{-t} \) is never zero, we can simplify this to:

\[ 1 - t = 0 \implies t = 1 \text{ second} \]

**Step 3: Substitute \( t = 1 \) back into the position function to find the distance from the origin.**

Now we substitute \( t = 1 \) into the position function \( x(t) \):

\[ x(1) = 20(1) e^{-1} = 20 e^{-1} \]

**Step 4: Calculate the value of \( x(1) \).**

Using the approximate value of \( e \approx 2.718 \):

\[ x(1) = \frac{20}{e} \approx \frac{20}{2.718} \approx 7.36 \text{ meters} \]

Thus, the distance of the electron from the origin when it momentarily stops is:

\[ \frac{20}{e} \text{ meters} \]

### Final Answer:

The distance of the electron from the origin when it momentarily stops is \( \frac{20}{e} \) meters.

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