The position of a particle moving a...
Question
The position of a particle moving along x-axis given by x=(−2t^3+3t^2+5)m. The acceleration of particle at the instant its velocity becomes zero is
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The position of a particle moving along x-axis given by ( x=left(-2 t^{3}+3 t^{2}+5right) m . ) The acceleration of particle at the instant its velocity becomes zero is
(1) ( 12 mathrm{m} / mathrm{s}^{2} )
(2) ( -12 mathrm{m} / mathrm{s}^{2} )
(3) ( -6 mathrm{m} / mathrm{s}^{2} )
(4) Zero

JEE/Engineering Exams
Physics
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The position of a particle moving along x-axis given by x=(−2t^3+3t^2+5)m. The acceleration of particle at the instant its velocity becomes zero is
Fullscreen

( quad x=-2 t^{3}+3 t^{2}+5 )
( begin{aligned} v=& frac{d x}{d t}=-6 t^{2}+6 t+0 &=-6 t^{2}+6 t=6 t(1-t) end{aligned} )
( v=0 quad ) at ( t=0 ) and ( t=1 )
( frac{d v}{d t}=frac{d}{d t}left(-6 t^{2}+6 tright) )
( =-12 t+6 )
( t=180, quad a=-12+6=-6 mathrm{m} / mathrm{s}^{2} )

Option (3)

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