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A particle moves along a straight line such that its displacement at any time t is given by S\mathrm{=t^3-6 t^2+3 t+4}. The velocity, when its acceleration is zero is -

Option: 1

-9 m/s


Option: 2

22 m/s


Option: 3

13 m/s


Option: 4

2 m/s


Answers (1)

best_answer

\mathrm{s=t^3-6 t^2+3 t+4}

\mathrm{\text { Velocity }(\text { dst })=3 t^2-12 t+3-0} ------------(1)

\mathrm{\text { Acceleration }\left(\frac{d^2 s}{d t^2}\right)=6 t-12}--------------------(2)

from equation, acceleration is zero

\mathrm{\begin{aligned} & \frac{d^2 s}{d t^2}=0 \\ & 6 t-12=0 \\ & t=2 \mathrm{sec} \end{aligned}}

Put 2 sec in equation (1) 

\mathrm{\begin{aligned} \operatorname{Velocity}\left(\frac{d s}{d t}\right) & =3(2)^2-12 \times 2+3 \\ & =12-24+3=-9 \mathrm{~m} / \mathrm{s} . \end{aligned}}

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HARSH KANKARIA

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