Question

The displacement $x$ of a particle varies with time according to the relation, $x=\frac{a}{b}(1-e^{-bt})$. Then the correct statements is/are

• A. at $t$ = $1/b$, the displacement of the particle in nearly $\frac{2a}{3b}$.
• 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 $x^{\prime }>\frac{a}{b}$.
• D. all of the above

Answer 1

Answer: (D)

all of the above

$x=\frac{a}{b}(1-e^{-bt})$
At $t$= $1/b$ , $x=\frac{a}{b}(1-\frac{1}{e})$

Therefore, option A is correct.

$1-\frac{1}{e}$ is approximately $2/3$. Hence $x=\frac{2}{3}a/b$
Maximum value of $x$ is $a/b$, given $t$ is positive.

Therefore, option C is correct.

Differentiating with t,
Velocity $u=\frac{dx}{dt}=\frac{a}{b}(+b{e}^{-bt})=a{e}^{-bt}$
$\Rightarrow u=a$ at $t=0$
Differentiating velocity with t,

Acceleration$=\frac{d^{2}x}{d{t}^2}=a{e}^{-bt}(-b)=-ab{e}^{-bt}$
$\Rightarrow$ acceleration $=-ab$ at $t=0$

Therefore, option B is correct.

Hence, correct option is D