The motion of a particle along a straight line is described by equation $x=10+27 t-t^{3}$ where $x$

Question

The motion of a particle along a straight line is described by equation $x=10+27 t-t^{3}$ where $x$ is in metre and $t$ in second. The retardation of the particle when its velocity becomes zero is

Professor PC Thomas Classes · Answer

(a) : Given : $x=10+27 t-t^{3}$ \ Velocity, $v=\frac{d x}{d t}=27-3 t^{2}$ \ When $v=0 \implies 27-3 t^{2}=0$ or $t=3 \mathrm{~s}$ \ $a=\frac{d v}{d t}=-6 t$ \ $a_{(t=3 \mathrm{~s})} = -6(3) = -18 \mathrm{~ms}^{-2}$ \ Retardation $= 18 \mathrm{~m} \mathrm{~s}^{-2}$