PYQ NEET- Motion In A Straight Line Kinematics L-8

Question: The motion of a particle along a straight line is described by equation $x=8+12 t-t^3$ where $x$ is in metre and $t$ in second. The retardation of the particle when its velocity becomes zero is#

A) $24 \mathrm{~m} \mathrm{~s}^{-2}$

B) zero

C) $6 \mathrm{~m} \mathrm{~s}^{-2}$

D) $12 \mathrm{~m} \mathrm{~s}^{-2}$

Answer: $12 \mathrm{~m} \mathrm{~s}^{-2}$#

Sol:#

Given $x=8+12 t-t^3$

Velocity, $\mathrm{v}=\frac{d x}{d t}=12-3 \mathrm{t}^2$

When $v=0$, then $12-3 t^2=0$ $$ \Rightarrow \mathrm{t}=2 \mathrm{~s} $$ $$ a=\frac{d v}{d t}=-6 \mathrm{t} $$ $\therefore \mathrm{At} \mathrm{t}=2 \mathrm{~s}, a=-12 \mathrm{~m} / \mathrm{s}^2$ $\therefore$ Retardation $=12 \mathrm{~m} / \mathrm{s}^2$