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: Write velocity as rate of change of displacement.
Given, \[ v=e^{-\beta x} \] Since, \[ v=\frac{dx}{dt} \] we get \[ \frac{dx}{dt}=e^{-\beta x} \]

Step 2: Separate the variables.
Taking \(x\)-terms to one side and \(t\)-terms to the other side, \[ e^{\beta x}dx=dt \]

Step 3: Integrate both sides.
Integrating, \[ \int e^{\beta x}dx=\int dt \] \[ \frac{1}{\beta}e^{\beta x}=t+C \]

Step 4: Apply initial condition.
At \[ t=0,\quad x=0 \] So, \[ \frac{1}{\beta}e^{0}=0+C \] \[ C=\frac{1}{\beta} \] Therefore, \[ \frac{1}{\beta}e^{\beta x}=t+\frac{1}{\beta} \] Multiplying by \(\beta\), \[ e^{\beta x}=1+\beta t \] Taking logarithm on both sides, \[ \beta x=\log(1+\beta t) \] Hence, \[ x=\frac{1}{\beta}\log(1+\beta t) \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{1}{\beta}\log(1+\beta t)} \]