The relation 3x=3t−6 describes the position of a particle in one direction where ‘x’ is in meters and t in sec . The displacement when velocity is zero, is……….

Given that  3x=3t−6..........(1)⇒3x=9t2+36−36t Differentiate on both sides with respect to time∴3dxdt=18t+0−36⇒3V=18t−36 If V=0 then 18t−36=0⇒18t=36⇒t=2S…………..(2) as v=6t-12   acceleration of the particle is a=dvdt =6 ms-2=constant at t=o velocity of the particle is  v=-12 ms-1  displacement of the body s=ut +at22 by the time t=2 sec velocity becomes zero s=-12 ×2+6×42=-12 m∴ from equation …….(1)At  t=0 the initial position of the particle  x1=12mAt t=2 S the final position of the partial  x2=0∴ Displacement  S=change in position=x2−x1=−12 m