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

At terminal speed `(dv)/(dt)=0rArr6-3v_(t)=0rArrv_(1)=2ms^(-1)`