The motion of a body is given by the equation  $\frac{d\upsilon }{dt}=6-3\upsilon ,$ where  $\upsilon$ is speed in m/s and  ‘t’ is time in seconds. If the body is at rest at  $t=0$  then     $\left[\text{One\hspace{0.17em}or more\hspace{0.17em}\hspace{0.17em}options}\left(\text{s}\right)\text{\hspace{0.17em}correct}\right]$

Re-write given expression as $\frac{d\upsilon }{2-\upsilon }=3dt.$  Given  $\upsilon =0$ at $t=0.$  Integrate both sides ${\int }_0^{\upsilon }\frac{d\upsilon }{2-\upsilon }=3{\int }_0^{t}dt,$ to get  $\upsilon =2(1-e^{-3t}).$ Thus, terminal velocity of the body is ${\upsilon }_t=\underset{t\to \infty }{\lim }2(1-e^{-3t})=2m/s.$ The acceleration of the body  is  $a=\frac{d\upsilon }{dt}=6-3\upsilon =6e^{-3t}$ which gives initial acceleration as  $6m/s^2$

A typical example of this motion is the motion of a raindrop Initially, the speed of the raindrop in-creases as it starts falling under gravity. As the speed increases, air resistance also increases which balances the gravitational pull after some time. Then  onwards the drop falls at a constant speed (terminal speed).