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

According to question, velocity of unit mass varies asv(x)=βx-2n                                          . . . .(i)dvdx=-2nβx-2n-1                        . . . .(ii)Acceleration of the particle is given by a=dvdt=dvdx×dxdt=dvdx×vUsing equation (i) and (ii), we get a=-2nβx-2n-1×βx-2n=-2nβ2x-4n-1