For a particle in uniform circular motion, the acceleration a⃗ at any point P(R,θ) on the circular path of radius R is (when θ is measured from the positive x-axis and v is uniform speed)

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(3) -(v²)/R cos θ î-(v²)/R sin θ ĵ. Idea: in uniform circular motion the acceleration points straight at the centre with magnitude (v²)/R. So write down the inward unit vector and scale it. The point P is at r⃗=R cos θ î+R sin θ ĵ, so the outward unit vector is r̂=cos θ î+sin θ ĵ and the inward one is -r̂. a⃗=(v²)/R(-r̂)=-(v²)/R cos θ î-(v²)/R sin θ ĵ. A quick check at θ=0: the particle is at (R,0) and the acceleration should point in the -î direction, back towards the origin. Only this option does that.

For a particle in uniform circular motion, the acceleration a⃗ at any point P(R,θ) on the circular path of radius R is (when θ is measured from the positive x-axis and v is uniform speed)

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Why the other options are wrong