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
B
12 m/s
C
42 m/s
D
9 m/s

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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