Question

The equation of motion of a particle is given by $s=2t^3-9t^2+12t+1$,where s and t are measured in cm and sec. The time when the particle stops momentarily is

• A. 1 sec
• B. 2 sec
• C. 1, 2 sec
• D. None of these

Answer 1

The correct option is C 1, 2 sec

$\frac{ds}{dt}=6t^2-18t+12$ =Velocity =0
(when particle stopped)
$\Rightarrow 6t^2-18t+12=0\Rightarrow (t-1)(t-2)=0$
Hence time 1, 2 sec.