Equation of motion of a body moving along x-axis at an instant t second is given by x = 40 +12t - t3. Displacement of the particle before coming to rest is:

a. 16 m

b. 56 m

c. 24 m

d. 40

find the velocity and time relation by differentiating the given equation w.r.t t.

By differentiating X we get the value of Velocity.i.e., V=dx/dt=d(40 +12t - t^3)/dt V=12-3t^2Now, before coming to rest velocity of the body is "zero" i.e., V=12-3t^2=0 12=3t^2 4=t^2 " t=2sec"Then, at "t=2sec" displacement(X™) of the body is X™=40+12(2)-(2)^3=56mAt initial position(when the time "t" is equal to "zero") displacement( X°) is given by X°=40+12(0)-(0)^3=40mHence displacement of the body before coming to rest is given by X=X™-X°=56-40Therefore X=16m.

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= 12-3t²at time t =t (the time when the particle comes to rest ):v= 0;=> 12-3t² = 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