A particle moves along a straight line O X . At a time t (in second) the distance x (in metre) of the particle from O is given by x=40+12 t-t^3 How long would the particle travel before coming to rest?

Step-by-step solution

Key Idea : Speed is rate of change of distance. Distance travelled by the particle is x=40+12 t-t^3 We know that, speed is rate of change of distance i e, v= d x d t aligned v & = d d t (40+12 t-t^3 ) & =0+12-3 t^2 aligned but final velocity v=0 12-3 t^2=0 or t^2= 12 3 =4 or t=2 ~s Hence, distance travelled by the particle before coming to rest is given by aligned x & =40+12(2)-(2)^3 & =40+24-8=64-8 & =56 ~m aligned