A particle moving with uniform retardation covers distances 18m, 14m & 10m in successive seconds. It comes to rest after travelling a further distance of

At time t=0     u is velocitygiven distance in successive seconds as 18m,14m,10mlet initial velocity =u; acceleration =a; distance covered in nth second isS=u+a22n−1put n=1→S1=u+a2(2×1-1)18=u+a2--→(1)S2=u+a22n−1put n=2→14=u+a22×2−114=u+32a--→(2)solving equations (1)&(2)u=20m/sa=-4m/s2by question,particle finally comes to rest v=0total distance travelled by the body  s=v2-u22a ⇒ s=02-2022(-4)s=50m=total distance travelled by the particleIt stops after travelling further distance after the three successive seconds =50-(18+14+10)=8m