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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
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The correct answer is:
$91 \mathrm{m}$
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, $t=3 \mathrm{s}$
$$
x=37+27 \times 3-(3)^{2}=91 \mathrm{m}
$$
$$
v=\frac{d x}{d t}=27-3 t^{2}
$$
According to problem,
$$
v=0 \Rightarrow 27-3 t^{2}=0
$$
Here, $t=3 \mathrm{s}$
$$
x=37+27 \times 3-(3)^{2}=91 \mathrm{m}
$$
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