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

A particle of unit mass undergoes one dimensional motion such that its

1.	A particle of unit mass undergoes one dimensional motion such that its velocity varies according to v(x) = = beta x ^(-2n)where B and n are constants and x is the position of the particle. The acceleraion of the particle as a function of x, is given by :
Answer» `-2n BETA ^(2) n ^(- 4n-1)` `- 2 beta ^(2) X^(-2 n+1)` `-2n beta ^(2) E^(-4n +1)` `-2n beta ^(2) x ^(-2n-1)` Solution :`v = beta x ^(-2n)` `So (DV )/(dx) = - 2n beta x ^(-2n -1)` Now `a = v (dv)/(dx) = (beta x ^(-2n)) (-2n beta x ^(-2n -1))` `implies a =- 2 n beta ^(2) x ^(-n -1)`