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
$x=40+12 t-t^{3}$
We know that, speed is rate of change of distance
$i e, v=\frac{d x}{d t}$
$\therefore v =\frac{d}{d t}\left(40+12 t-t^{3}\right) $
$=0+12-3 t^{2}$
but final velocity $v=0$
$\therefore 12-3 t^{2}=0 $
or $ t^{2}=\frac{12}{3}=4$
or $ 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 $
$=56 \,m$