A particle moves on a circular path of radius R with constant tangential acceleration a_t. If it starts from rest, what is the angle between its velocity vector and total acceleration vector when it has completed one full revolution?

Answer and explanation

Correct answer: D. tan^(-1)(4π)

  1. Step 1: After one full revolution, distance s = 2π R.
  2. Step 2: v^2 = 2 a_t s = 2 a_t (2π R) = 4π R a_t.
  3. Step 3: Centripetal acceleration a_c = v^2 / R = 4π a_t.
  4. Step 4: The angle θ between velocity (which is purely tangential) and total acceleration is: tan θ = a_c / a_t = (4π a_t) / a_t = 4π => θ = tan^(-1)(4π).

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