PhysicsMedium42×since 2002Q6262For a particle in uniform circular motion the acceleration a→\overrightarrow aa at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis)A−v2Rcosθi^+v2Rsinθj^- {{{v^2}} \over R}\cos \theta \widehat i + {{{v^2}} \over R}\sin \theta \widehat j−Rv2cosθi+Rv2sinθjB−v2Rsinθi^+v2Rcosθj^- {{{v^2}} \over R}\sin \theta \widehat i + {{{v^2}} \over R}\cos \theta \widehat j−Rv2sinθi+Rv2cosθjC−v2Rcosθi^−v2Rsinθj^- {{{v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j−Rv2cosθi−Rv2sinθjDv2Ri^+v2Rj^{{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat jRv2i+Rv2jCheck answerSkip