Q: 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 (1/8) of the circumference of the circle is :

(a) Zero

(b) 0.500 v

(c) 0.765 v

(d) 0.125 v

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Ans: (c)

Sol: Change in velocity , $\displaystyle \vec{\Delta v} = \vec{v_2} – \vec{v_1}$

$\displaystyle |\vec{\Delta v} |= \sqrt{v^2 + v^2 – 2 v \times v \times cos\theta } $ (since v_1 = v_2 = v and θ = 2π/8)

$\displaystyle |\vec{\Delta v} |= \sqrt{2 v^2 (1 – cos\theta )} $

$\displaystyle |\vec{\Delta v} | = \sqrt{2 v^2 (2 sin^2(\frac{\theta}{2}))} $

$\displaystyle |\vec{\Delta v} | = 2 v sin(\frac{\theta}{2}) $

$\displaystyle |\vec{\Delta v} | = 2 v sin(\frac{\pi}{8}) $

= 0.765 v