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

Explanation

(a) `(dv)/(dt)=6-3v=3(2-v)`
`int_(0)^(v)(dv)/((2-v))=3int_(0)^(t) dt`
`(| log_(e)(2-v)|_(0)^(v))/-1=3|t|_(0)^(t)`
`log_(e)(2-v)-log_(e)v=-3t`
`log_(e)((2-v)/v)=-3t`
`(2-v)/v=e^(-3t)`
`2-v=2e^(-3t)`
`v=2(1-e^(-3t))`
(b) Terminal velocity, i.e. at `t=oo`,
`v=2(1-e^(-3t))=2(1-e^(-oo))` ,
`=2(1-1/e^(oo))=2(1-1/oo)`
`=2(1-0)=2m//s`