Question

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

• A. $24\, m\, s^{-2}$
• B. zero
• C. $6\, m\, s^{-2}$
• D. $12\, m\, s^{-2}$

Answer 1

Answer: (D)

$12\, m\, s^{-2}$

Answer 2

$x=8+12 t-t^{3}$
$v=\frac{d x}{d t}=12-3 t^{2}$
$a=\frac{d v}{d t}=-6 t$
putting $v =0$
$3 t ^{2}=12$
$t =2 sec$
$\therefore a =-6 \times 2=-12 ms ^{-2}$
$\therefore$ retardation $=12 ms ^{-2}$