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Question

A particle is travelling along a straight line $$OX$$. The distance $$x$$ (in metre) of the particle from $$O$$ at a time $$'t'$$ is given by $$x = 37 + 27t - t^{3}$$, where 't' is time in second. The distance of the particle from $$O$$ ,when it comes to rest is:


A
81 m
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B
91 m
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C
101 m
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D
111 m
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Solution

The correct option is C $$91\ m$$
The position of particle is given as $$x = 37 + 27 t - t^{3}$$
Thus the velocity of particle  $$v = \dfrac {dx}{dt} = 27 - 3t^{2}$$
According to problem,
$$v = 0\Rightarrow 27 - 3t^{2} = 0$$
Here, we get $$t = \sqrt{\dfrac{27}{3}} = \sqrt{9}=3s$$
Thus the distance of particle when it comes to rest   $$x (t=3 \ s) = 37 + 27 \times 3 - (3)^{3} = 91\ m$$

Physics

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