A particle moves along a straight line O X . At a time t (in...

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=\frac{d x}{d t}$$\begin{aligned} \therefore \quad v & =\frac{d}{d t}\left(40+12 t-t^3\right) \\ & =0+12-3 t^2\end{aligned}$but final velocity $v=0 \therefore \quad 12-3 t^2=0$ or $\quad t^2=\frac{12}{3}=4 or t=2 \mathrm{~s}$ Hence, distance travelled by the particle before coming to rest is given by\begin{aligned} x & =40+12(2)-(2)^3 \\ & =40+24-8=64-8 \\ & =56 \mathrm{~m}\end{aligned}$$