Question

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

• A. $24m/s^2$
• B. zero
• C. $6m/s^2$
• D. $12m/s^2$

Answer 1

The correct option is D $12m/s^2$

$x=8+12t-t^{3}v=\frac{dx}{dt}=12-3t^2$
When v=0
$12-3t^2=0$
t=2sec. [at (t = 2s) velocity becomes zero]
$a(t=2)=\frac{dv}{dt}\Rightarrow \frac{d}{dt}[12-3t^2]=0-6t=-6ta=-6\times 2$
$a=-12m/s^2$ (-ve sign implies retardation)