=(3t2−18t+6)ms−1 So, the particle comes to rest since this equation has two valid roots. ⇒t1=(3−√7)s and ⇒t2=(3−√7)s Substituting the value t=2s in the equation for velocity, we get, v=−18ms−1 The acceleration of the particle is a =6t−18ms−2 Substituting the value t=2s in the equation for acceleration, we get, a=6ms−2