Question

A particle is moving along x-axis with velocity \(v=e^{-\beta x}\). At time \(t=0\), the particle is located at \(x=0\). The displacement of the particle as a function of time is:

• A. \(e^{-\beta t}\)
• B. \(\dfrac{1}{\beta}e^{(1-\beta t)}\)
• C. \(\dfrac{1}{\beta}\log[1-\beta t]\)
• D. \(\dfrac{1}{\beta}\log[1+\beta t]\)

Hint:

When velocity is given as a function of position, use \(v=\frac{dx}{dt}\), separate variables, and then apply the initial condition.

Answer 1

The Correct Option is D

Step 1: Separate variables and integrate.
\( \frac{dx}{dt} = e^{-\beta x} \Rightarrow e^{\beta x}dx = dt \). Integrating both sides: \( \frac{e^{\beta x}}{\beta} = t + C \). At \( t = 0, x = 0 \): \( C = \frac{1}{\beta} \).

Step 2: Solve for x(t).
\( e^{\beta x} = 1+\beta t \Rightarrow \beta x = \ln(1+\beta t) \). \[ \boxed{x = \dfrac{1}{\beta}\log(1+\beta t)} \]