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Question

A particle moves along a straight line such that its displacement at any time $$t$$ is given by $$s = t^3 - 6t^2 + 3t  + 4$$ metres. The velocity when the acceleration is zero is:-


A
3 m/s
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B
12 m/s
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C
42 m/s
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D
9 m/s
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Solution

The correct option is D $$-9 \ m/s$$
$$S =t^3 - 6^2 +3t +4$$
$$v= \dfrac{ds}{dt} $$

So $$= 3t^2 -12 +3 =v $$
again differentiate ,$$a = \dfrac{dv}{dt}$$
$$6t -12 = 0$$
$$t = 2 s$$
$$v = 3(2)^2 - 12 \times 2 +3 = -9ms^{-1}$$

Hence (D) is correct answer.

Physics

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