Question

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

Answer 1

Answer: (C)

Key Idea Speed is rate of change of distance. Distance travelled by the particle is
$x=40+12t-t^3$
We know that, speed is rate of change of distance
$ie,v=\frac{dx}{dt}$
$\therefore v=\frac{d}{dt}\left(40+12t-t^3\right)$
$=0+12-3t^2$
but final velocity $v=0$
$\therefore 12-3t^2=0$
or $t^2=\frac{12}{3}=4$
or $t=2s$
Hence, distance travelled by the particle before coming to rest is given by
$x=40+12(2)-(2{)}^3$
$=40+24-8=64-8$
$=56m$