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Mechanical Engineering QUESTION #2531
Question 1
For a particle moving in a circular path of radius \(R\) with constant speed \(v\), the magnitude of total acceleration is:
  • \(\frac{v^2}{R}\) directed toward the center✔️
  • \(\frac{v}{R}\) tangentially
  • \(\dot{v}\) only
  • \(\sqrt{\dot{v}^2 + (v^2/R)^2}\) — always larger than \(v^2/R\)
Correct Answer Explanation
For constant speed circular motion, \(\dot{v}=0\), so the tangential acceleration is zero. Only the normal (centripetal) acceleration exists: \(a_n=v^2/R\), directed toward the center.