Question

A particle is moving along x-axis with velocity \( v = \frac{1}{1 + \beta t} \). At time \( t = 0 \), the particle is at \( x = 0 \). The displacement as a function of time is:

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

Hint:

Displacement from Velocity}
Displacement is the integral of velocity.
Know standard integrals like \( \int \frac{1}{a + bt} dt = \frac{1}{b} \log(a + bt) \)
Always apply initial conditions to evaluate constants.

Answer 1

The Correct Option is D

Step 1: Given \( v = \frac{dx}{dt} = \frac{1}{1 + \beta t} \) Step 2: Integrate to find \( x(t) \) \[ \int dx = \int \frac{1}{1 + \beta t} dt \Rightarrow x = \frac{1}{\beta} \log(1 + \beta t) + C \] Step 3: Apply initial condition: At \( t = 0 \), \( x = 0 \Rightarrow C = 0 \) Final Result: \( \boxed{x(t) = \frac{1}{\beta} \log(1 + \beta t)} \)