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A particle moves along a straight line OX. At a time t (in second) the distance x (in metre) of the particle from O is given by $x = 40 + 12t - t^{3}$. How long would the particle travel before coming to rest?
Explanation
When particle comes to rest, $V = 0 = dx/dt = d/dt(40 + 12t - t^{3})$ $\Rightarrow12 - 3t^{2} = 0 \Rightarrow t^{2} = 12/3 = 4 \therefore t = 2\text{sec}$ Therefore distance travelled by particle before coming to rest, $x = 40 + 12t - t^{3} = 40 + 12 \times 2 - (2)^{3} = 56m$