Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A particle moves along a straight line such that its displacement at any time t is given by <img alt="\mathrm{s=\left(t^3-6 t^2+3 t+4\right) \mathrm{m}.}" src="https://entrancecorner.oncodecog

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

Option: 1

\mathrm{-9 \mathrm{~m} / \mathrm{s}^2}


Option: 2

\mathrm{-10 \mathrm{~m} / \mathrm{s}^2}


Option: 3

\mathrm{-6 \mathrm{~m} / \mathrm{s}^2}


Option: 4

\mathrm{-4 \mathrm{~m} / \mathrm{s}^2}


Answers (1)

best_answer

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

\mathrm{\text { velocity } v=\frac{d s}{d t}=\frac{d}{d t}\left[t^3-6 t^2+3 t+4\right]}

                                  \mathrm{=3 t^2-12 t+3\, \, \, \, \, \, -(1)}

\mathrm{\text { Acceleration } a=\frac{d v}{d t}=\frac{d}{d t}\left[3 t^2-12 t+3\right]}

                                      \mathrm{a=6 t-12 \text {\, \, \, \, - (2) }}

but acceleration is zero-
                                            \mathrm{ a=\frac{d v}{d t}=0 }

\mathrm{from (2) -}

                                         \mathrm{\begin{array}{r} 0=6 t-12 \\ t=2 \mathrm{sec} \end{array}}

Now, 

          put\mathrm{ t=2\, \mathrm{sec} } in equation - (1) we get velocity 

\mathrm{ \begin{aligned} & v=3(2)^2+12 \times 2+3 \\ & v=12-24+3 \\ & v=-12+3=-9 \mathrm{~m} / \mathrm{s} \text { Ans. } \end{aligned}}

Posted by

qnaprep

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

'PM Modi personally supervising the situation': Centre tells SC in NEET case
anam-khan

NEET Paper Leak Row: NTA informs SC of security overhaul, institutional reforms
anam-khan

Re-NEET 2026 City Intimation Slip LIVE: NTA MBBS admit card soon; direct link, login details, exam date

Latest Question