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

Question

A particle moves along a straight line. At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from the origin is given by $x=20+27 t-t^{3}$. How long would the particle travel before coming to rest?

Professor PC Thomas Classes · Answer

(b) : Position $x=20+27 t-t^{3}$ \ Velocity, $v=\frac{d x}{d t}=27-3 t^{2}$ \ When particle comes to rest, $v=0 \implies 27-3 t^{2}=0 \implies t=3 \mathrm{sec}$. \ Initial position at $t=0$ is $x_{0} = 20 \mathrm{~m}$. \ Final position at $t=3\mathrm{s}$ is $x_{3} = 20 + 27(3) - (3)^{3} = 20 + 81 - 27 = 74 \mathrm{~m}$. \ Distance travelled $= x_{3} - x_{0} = 74 - 20 = 54 \mathrm{~m}$