Answer is: option2
\( \frac{1}{e^2} \)Solution:
A particle is at rest when its velocity is zero.
Given position function:
\[ x(t) = \sqrt{x} \ln t \]
Velocity is the derivative of position:
\[ V(t) = \sqrt{x} \left(\frac{1}{x}\right) + \ln x \cdot \frac{1}{2\sqrt{x}} \]
Simplifying:
\[ V(t) = \frac{1}{\sqrt{x}} + \frac{\ln x}{2\sqrt{x}} \]
Factoring:
\[ V(t) = \frac{1}{\sqrt{x}} \left( 1 + \frac{\ln x}{2} \right) = 0 \]
Setting the term in parentheses to zero:
\[ 1 + \frac{\ln x}{2} = 0 \]
\[ \ln t = -2 \]
Solving for \( t \):
\[ t = e^{-2} = \frac{1}{e^2} \]
Final : \( t = \frac{1}{e^2} \)