A particle of unit mass undergoes $$one-dimensional$$ motion such that its velocity varies according $$tov(x)=$$β$$x-2nwhere$$ β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

Question

A particle of unit mass undergoes $$one-dimensional$$ motion such that its velocity varies according $$tov(x)=$$β$$x-2nwhere$$ β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

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According to question, velocity of unit mass varies $$asv(x)=$$β$$x-2n$$                                          . . . .(i)dvdx=-2n$$β$$x-2n-1$$                        . . . .$$(ii)Acceleration$$ of the particle is given by $$a=dvdt=dvdx$$×$$dxdt=dvdx$$×vUsing equation $$(i) and (ii), we get $$a=-2n$$β$$x-2n-1$$×β$$x-2n=-2n$$β$$2x-4n-1$$

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to
$v (x) = β x^{- 2 n}$
where $β$ and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

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A particle of unit mass undergoes one-dimensional motion such that its velocity varies according tov(x)=βx-2nwhere β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

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Ready to Test Your Skills?

free study material

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to
$v (x) = β x^{- 2 n}$
where $β$ and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by