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 statements is correct?

• A. At $t=\frac{1}{b}$, the displacement of the particle is $\frac{a}{b}$.
• B. The velocity and acceleration of the particle at t = 0 are a and b respectively.
• C. The particle will come back to its starting point as $t\to \infty$.
• D. None of these

Answer 1

The correct option is D

None of these

Given that, $x=\frac{a}{b}(1-e^{-bt})$
Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}\left\{\frac{a}{b}\right(1-e^{-bt}\left)\right\}=a{e}^{-bt}$
Acceleration, $\alpha =\frac{dv}{dt}=\frac{d}{dt}\left(a{e}^{-bt}\right)=-ab{e}^{-bt}$
At $t=\frac{1}{b}$, the displacement of the particle is
$x=\frac{a}{b}(1-e^{-1})\simeq \frac{a}{b}(1-\frac{1}{3})\simeq \frac{2}{3}\frac{a}{b}(\because e^{-1}\simeq \frac{1}{3})$
Hence, option (a) is incorrect.
At t = 0, $x=0,v=a{e}^{-0}=aand\alpha =-ab{e}^{-0}=-ab$.
Hence, option (b) is also incorrect.
When $t\to \infty$, $x=\frac{a}{b}(1-e^{-\infty })=\frac{a}{b}$. Hence, option (c) is also incorrect.
Thus, the correct option is (d).