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 sec. The retardation of the particle when its velocity becomes zero is :

• A. $24{ms}^{-2}$
• B. $zero$
• C. $6{ms}^{-2}$
• D. $12{ms}^{-2}$

Answer 1

Answer: (D)

$12{ms}^{-2}$

$x=8+12t-t^{3}V=\frac{dx}{dt}=12-3t^2$

$\therefore V=0$ at $t=\pm 2s$, negative time has no significance,

$\therefore t=2s$

$a=\frac{dV}{dt}=-6t\therefore a$ at $t=2=-6\times 2=-12m/s^2$

Thus, retardation when velocity becomes $0$ is $12m{s}^{-2}$

So, option $D$ is correct.