A particle moves along a straight line OX . At a time t (in seconds) the distance x (in metres) 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

The given equation is x=40+127-t^3 Differentiating we get u= d x d t =12-3 t^2 When particle comes to rest means array l & v = 0 & 12-3 t^2 =0 & t =2 sec . array The distance travelled before coming to rest aligned ₀^5 d s & = ₀^2 v d t 5 & = ₀^2 (12-3 t^2 ) d t & = [12 t- 3 t^3 3 ]₀^2 & =12 2-(8-0) & =16 ~m . aligned