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Question

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x)=βx2n ,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 :


  1. -2nβ2e-4n+1

  2. -2nβ2x-2n-1

  3. -2nβ2x-4n-1

  4. -2nβ2x-2n+1


Solution

The correct option is C

-2nβ2x-4n-1


v(x)=βx2na=dvdt=dvdx.dxdt=dvdx.vdvdx=2nβx2n1a=2nβx2n1.βx2n   =2nβ2x4n1

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