-->

Motion In A Plane

Question
CBSEENPH11020526

A particle of unit mass undergoes one-dimensional motion such that its velocity according to
V(x) =  βx-2n 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 

  • -2nβ2 x-2n-1

  • -2nβx-4n-1

  • -2β x-2n+1

  • -2nβ2e-4n+1

Solution

B.

-2nβx-4n-1

Given, v = βx-2n
straight a equals space dv over dt equals dx over dt. dv over dx
straight a equals straight v dv over dx equals left parenthesis space βx to the power of negative 2 straight n end exponent right parenthesis left parenthesis negative 2 straight n space βx to the power of negative 2 straight n minus 1 end exponent right parenthesis
straight a equals negative 2 nβ squared straight x to the power of negative 4 straight n minus 1 end exponent