Question

The motion of a body is given by the equation $\frac{dv}{dt}=6-3v:$where $v$ is in $m/s$. If the body was at rest at $t=0$
(i) the terminal speed is $2m/s$
(ii) the magnitude of the initial acceleration is $6m/s$
(iii) the speed varies with time as $v=2(1-e^{-3t})m/s$
(iv) The speed is $1m/s$, when the acceleration is half the initial

• A. $\left(i\right),\left(iii\right)$
• B. $\left(ii\right),\left(iii\right),\left(iv\right)$
• C. $\left(i\right),\left(ii\right),\left(iii\right)$
• D. $All$

Answer 1

The correct option is A $\left(i\right),\left(iii\right)$

$\frac{dv}{dt}=6-3v$

$\int \frac{dv}{6-3v}=\int dt$

$-\frac{1}{3}\log |2-v|=dt$

$-\log |2-v|=3t$

$|2-v|=e^{-3t}$

$2-e^{-3t}=v$

Initial acceleration is $0$.

At $t=0$, terminal velocity is $v=2m/s$

Hence, condition 1 and 3 are correct.