Question

For a body moving along $x-$axis, its velocity is given by $v=\frac{a}{b}(1-e^{-bt})$ where $a$ and $b$ are positive constant. At $t=0,x=0$. Position $x$ of the particle with time $t$ is given by

• A. $\frac{a}{b}(1-e^{-bt})t$
• B. $\frac{a}{b}[t+\frac{e^{-bt}}{b}]$
• C. $\frac{a}{b}[t+\frac{1}{b}(e^{-bt}-1\left)\right]$
• D. $\frac{a}{b}[t-\frac{1}{b}(e^{-bt}-1\left)\right]$

Answer 1

The correct option is C $\frac{a}{b}[t+\frac{1}{b}(e^{-bt}-1\left)\right]$

$V=\frac{dx}{dt}=\frac{a}{b}(1-e^{-bt})$

$\int dx=\int \frac{a}{b}(1-e^{-bt})$

$x=\frac{a}{b}[t+\frac{e^{-bt}}{b}]+c$

$x=0$ at $t=0$, putting

$c=-\frac{a}{b^2}$

$x=\frac{a}{b}[t+\frac{1}{b}(e^{-bt}-1\left)\right]$