A particle leaves O O\,O at time t=0t=0t=0 with velocity v=(3t+5−2t2) m s−1v=(3t+5-2t^2)\text{ m s}^{-1}v=(3t+5−2t2) m s−1.
Find the value of t t\,t and the speed of the particle when it stops accelerating.
Find the distance of the particle from O O\,O when it changes direction.