Question

The motion of a body is given by the equation $\frac{dv}{dt}=6.0-3v$, where $v$ is the speed (in $m{s}^{-1}$ ) and $t$ is time in second. If the body was at rest at $t=0$, then find its speed (in $m/s$ ) when the acceleration is half the initial value.

Answer 1

Answer: 1

$a=\frac{dv}{dt}=(6-3v)$
$\int \frac{dv}{6-3v}=\int dt$
${\frac{\ln (6-3v)}{-3}|}_0^v=t$
$-\frac{1}{3}\left[\ln \left\{\frac{(6-3v)}{(6)}\right\}\right]=t$
$-\frac{1}{3}\left[\ln \left\{\frac{(2-v)}{(2)}\right\}\right]=t$
$v=2\left(1-e^{-3t}\right)$
$a=\frac{dv}{dt}=6e^{-3t}$
$3=6e^{-3t}$
$e^{-3t}=\frac{1}{2}$
$v=2\left[1-\frac{1}{2}\right]=1m/s$