A particle moves along a circle with constant speed $2 m s^{- 1}$ . If it completes $1 / 4 t h$ of rotation, change in its velocity as

2 m s^{- 1}

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\sqrt{2} m s^{- 1}

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1 / 2 m s^{- 1}

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2 \sqrt{2} m s^{- 1}

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Solution

The correct option is D $2 \sqrt{2} m s^{- 1}$
after completing 1/4th of the circle the angle between initial and final velocities is $90^{\circ}$

so $|_{2} -_{1} | = \sqrt{v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} c o s θ}$

= $\sqrt{2^{2} + 2^{2}}$ m/s

= $2 \sqrt{2}$ m/s which is the change in the velocity.

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A particle moves along a circle with constant speed 2ms−1. If it completes 1/4th of rotation, change in its velocity as

The correct option is D 2√2ms−1after completing 1/4th of the circle the angle between initial and final velocities is 90∘ so |→v2−→v1|=√v21+v22−2v1v2cosθ = √22+22 m/s = 2√2 m/s which is the change in the velocity.

A particle moves along a circle with constant speed $2 m s^{- 1}$ . If it completes $1 / 4 t h$ of rotation, change in its velocity as