A particle moves along the side AB, BC, CD of a square of side $25 m$ with a velocity of $15 m / s$ . Its average velocity is

15 m / s

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10 m / s

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7.5 m / s

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5 m / s

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Solution

The correct option is D $5 m / s$
As shown in the figure, total distance will be of length $3$ sides but total displacement will be just side $A D$ .
Total time $= \frac{25 + 25 + 25}{15} = 5 s$
Total displacement $= 25 m$
Average velocity $= \frac{25}{5} = 5 m / s$

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