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 + 12t - t³

How long would the particle travel before coming to rest ?

56 m

\(Displacement : x = 40 + 12t - t^3\)

\(Velocity : v = \frac{ds}{dt} = 12 - 3t^2\)

\(Acceleration : a = \frac{dv}{dt} = - 6t\)

particle will come to rest when \(v = 12 - 3t^2 = 0 ⇒ t = 2 s\)

x = 40 +12(2) -(2)³

x = 56 m