Question

The motion of a body is given by the equation $\frac{dv}{dt}=6-3v$ where $v$ is the speed in $m{s}^{-1}$ and $t$ is time in $s$ . The body is at rest at $t=0$. The speed varies with time as

• A. $v=(1-e^{-3t})$
• B. $v=2(1-e^{-3t})$
• C. $v=(1+e^{-2t})$
• D. $v=2(1+e^{-2t})$

Answer 1

Answer: (B)

$\frac{dv}{dt}=6-3v$ or $dt=\frac{dv}{6-3v}$
Integrating both sides, we get
$t=-\frac{1}{3}ln\left(6-3v\right)+C$
where $C$ is a constant of integration
At $t=0$, $v=0$
$\therefore C=\frac{1}{3}ln6$
$\therefore t=-\frac{1}{3}ln\left(\frac{6-3v}{6}\right)$ or $e^{-3t}=1-\frac{1}{2}v$ or $v=2\left(1-e^{-3t}\right)$