Q126 / 200 · JIPMER 2008 · Physics
A particle moves along a straight line $O X$. At a time $t$ (in second) the distance $x$ (in metre) of the particle from $O$ is given by
$x=40+12 t-t^3$
How long would the particle travel before coming to rest?
- A$24 m$
- B$40 m$
- Correct answerC$56 m$
- D$16 m$
Solution
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 $i e, v=\frac{d x}{d t}$
$\begin{aligned} \therefore \quad v & =\frac{d}{d t}\left(40+12 t-t^3\right) \\ & =0+12-3 t^2\end{aligned}$
but final velocity $v=0$
$\therefore \quad 12-3 t^2=0$
or $\quad t^2=\frac{12}{3}=4$
or $t=2 \mathrm{~s}$
Hence, distance travelled by the particle before coming to rest is given by
$\begin{aligned} x & =40+12(2)-(2)^3 \\ & =40+24-8=64-8 \\ & =56 \mathrm{~m}\end{aligned}$
Hence, the correct answer is option: 'C'.