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A particle’s displacement is given by $x=\dfrac{a}{b}(1-e^{-bt})$. Identify the incorrect statement.

1. When $t=1/b$, the displacement is approximately $2/3$ of $a/b$
2. At $t=0$, the particle’s velocity is $a$ and its acceleration is $-ab$
3. The particle cannot move beyond $x=a/b$
4. As $t\to\infty$, the particle does not return to its initial position

Correct answer: At $t=0$, the particle’s velocity is $a$ and its acceleration is $-ab$

Solution

From $x=\dfrac{a}{b}(1-e^{-bt})$, velocity is $v=\dfrac{dx}{dt}=ae^{-bt}$, so at $t=0$, $v=a$. Acceleration is $a_x=\dfrac{dv}{dt}=-abe^{-bt}$, so at $t=0$, it is $-ab$; this statement is actually correct, so the question’s intended incorrect statement is inconsistent with the provided answer key.

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