Question

The displacement $x$ of a particle varies with time according to the relation $x=\frac{a}{b}(1-e^{-bt})$. Which of the following is not correct? (a and b are the positive constants.)

• A. At $t=\frac{1}{b}$, acceleration of the particle is $-\frac{ab}{e}$.
• B. The velocity and acceleration of the particle at $t=0$ are $a$ and $-ab$, respectively.
• C. The particle cannot reach a point at a distance $x^{\prime }$ from its starting position if $x^{\prime }>a/b$.
• D. The particle will come back to its starting point as $t\to \infty$.

Answer 1

The correct option is D The particle will come back to its starting point as $t\to \infty$.

$x=\frac{a}{b}(1-e^{-bt})v=\frac{dx}{dt}=a{e}^{-bt}a=\frac{dv}{dt}=-ab{e}^{-bt}$

Now, At $t=\frac{1}{b}$
acceleration $=-\frac{ab}{e}$
velocity at $t=0$ is $a$
acceleration at $t=0$ is $-ab$
the particle can never come back to starting point because the term $(1-e^{-bt})$ is always less than $1$ for all $t$