Q. A particle moves along a straight line $OX.$ At a time $t$ (in second) the distance $x$ (in metre) of the particle from $O$ is given by $x = 40 + 12t - t^{3}.$ How long would the particle travel before coming to rest?
A) 24 m
B) 40 m
C) 56 m
D) 16 m
Explanation
(c) $v = \frac{dx}{dt} = 0 + 12t - 3t^{2} = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = 2\ s$
Hence, distance travelled by the particle before coming to rest is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 40 + 12(2) - (2)^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 40 + 24 - 8 = 64 - 8$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 56\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.