Question

A particle is moving in a circle of radius r with a constant speed V. The change in velocity after the particle has travelled a distance equal to $ \left( \frac{1}{8} \right) $ of the circumference of the circle is:

• A. zero
• B. 0.500 v
• C. 0.765 v
• D. 0.125 v

Answer 1

Answer: (C)

Angle traversed by the particle in $ \left( \frac{1}{8} \right) $ of the circumference of the circle $ =\left( \frac{1}{8} \right)\frac{2\pi r}{r} $ $ =\frac{\pi }{4}={{45}^{0}} $ So, change in velocity, ___ $ \Delta v=\sqrt{{{v}^{2}}+{{v}^{2}}-2{{v}^{2}}\,\cos ({{45}^{o}})} $ $ =\sqrt{2{{v}^{2}}-2{{v}^{2}}\times \frac{1}{\sqrt{2}}} $ $ =v\sqrt{2-\sqrt{2}} $ $ =0.765\,v $