At \\\\(t=0 ,\\\\) particle is at \\\\(,\\\\) let's say \\\\(x\\\\) distance ,from \\\\(O ;\\\\)then putting \\\\(t=0\\\\) in the given displacement-time equation we get\\\\(;\\\\)\\\\(x =40 +12(0) -(0)^3 = 40 m \\\\)Particle comes to rest that means velocity of particle becomes zero after travelling certain displacement \\\\(;\\\\) let's say the time be \\\\(t.\\\\)then after differentiating the given displacement\\\\(-\\\\)time equation \\\\(wrt.\\\\) time we get velocity\\\\(-\\\\)time equation\\\\(v= 12-3t^2\\\\)at time \\\\(t =t\\\\) \\\\((\\\\)the time when the particle comes to rest \\\\():\\\\)\\\\(v= 0;\\\\)\\\\(=> 12-3t^2 = 0;\\\\)\\\\(=> t = 2 s\\\\)Then \\\\(,\\\\)at \\\\(t =2s\\\\) we are at \\\\(,\\\\) let's say \\\\(x'\\\\) distance from \\\\(O ;\\\\)put this value of\\\\( t (=2)\\\\) in given displacement-time equation , we get\\\\(;\\\\)\\\\(x'= 40 +12(2) -(2)^3;\\\\)\\\\(= 56m\\\\)Further\\\\(;\\\\)We have seen that the particle started his journey when it is at \\\\(40m\\\\) from the point \\\\(O.\\\\) And came to rest at \\\\(56m\\\\) from the point \\\\(O .\\\\)then the particle traveled a distance of\\\\(:\\\\)\\\\(56-40=16m.\\\\)Hence,option \\\\((A)\\\\) is correct answer.