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EEJ MAIN Physics QUESTION #7094
Question 1
A particle moves in a circle at a constant $10\text{ ms}^{-1}$. What is the magnitude of velocity change after turning $60^{\circ}$?
  • $10\sqrt{3}\text{ m/s}$
  • zero
  • $10\sqrt{2}\text{ m/s}$
  • $10\text{ m/s}$✔️
Correct Answer Explanation

The magnitude of change in velocity is $|\Delta v| = 2v \sin(\frac{\theta}{2})$.

  • $|\Delta v| = 2(10) \sin(60^{\circ}/2) = 20 \sin 30^{\circ}$
  • $|\Delta v| = 20 \times \frac{1}{2} = 10\text{ m/s}$.