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

20 m 20e m $\frac{20}{e}m$ zero

$\frac{20}{e}m$

The correct answer is Option (3) → $\frac{20}{e}m$ $x = 20t e^{–t}$ $∴v=\frac{dx}{dt}=20\left(t\frac{de^{-t}}{dt}+e^{-t}×1\right)$ or $0=20[te^{-t}×(-1)+e^{-t}]$ $∴t = 1$ Thus $x = 20×1×e^{-1}=\frac{20}{e}$