are given a = 6(t- 1) Here, s = +x, v = +v. To find velocity-time relation using a = dv/dt, we have dtdv = 6 ( t - 1 )...(i)When velocity changes from v 0 to v during time t, integrating both sides with respect to time, we have∫ v 0 v dv=∫ 0t 6(t−1)dt⇒v−v 0 =3t 2 −6t Substituting v 0 = 2 m/s, we have v = dtdx =3t 2 −6t+2 (II)dx = (3t 2 −6t+2)dt.(ii) find position-time relation, again integrating (ii) both sides, we have ∫ 0x dx=∫ 0t (3t 2 −6t+2)dt This gives x=t 3 −3t 2 + 2t = t (t - 1)(t - 2) III)The particle passes through the origin where x = 0 from (iii), we have t = 0, 1s and 2s. That means the particle crosses the origin twice at t = 1 s and t = 2 s. After t = 2 s, x is positive. Hence, its displacement points in positive x-direction and goes on increasing its magnitude.The particle will COME at rest when v = 0. From (ii), we have t = (1− 3 1 )s (=t 1 , say and t = (1+ 3 1 )s (=t 2 , say)Hence, the particle at t =t 1 and t 2 . That signifies to and fro motion of the particle during first two seconds as it starts RETRACING its path at t = t 1 and t = t 2 The displacement (x), velocity (v), and acceleration (a) of the particle in different time intervals are given in the following table and SHOWN in the following graphs.Time interval (1− 3 1 )(1− 3 1 ) (1+ 3 1 )