Q138 / 160 · WBJEE 2012 · Physics
A particle is travelling along a straight line $O X$. The distance $x$ (in metre) of the particle from 0 at a time t is given by $x=37+27 t-t^{3}$, where t is time in seconds. The distance of the particle from $O$ when it comes to rest is
- A$81 \mathrm{m}$
- Correct answerB$91 \mathrm{m}$
- C$101 \mathrm{m}$
- D$111 \mathrm{m}$
Solution
Given, $x=37+27 t-t^{3}$
$$
v=\frac{d x}{d t}=27-3 t^{2}
$$
According to problem,
$$
v=0 \Rightarrow 27-3 t^{2}=0
$$
Here, $t=3 \mathrm{s}$
$$
x=37+27 \times 3-(3)^{2}=91 \mathrm{m}
$$
Hence, the correct answer is option: 'B'.