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Q.

The position vector of particle changes from 10 m east to 10 m north in one second, What is the average velocity of the Particle ?

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a

20 m/s

b

(10/2)m/s

c

5 m/s

d

102m/s

answer is C.

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Detailed Solution

The net displacement will be
Δx=(10j^10i^)m|Δx|=(10)2+(10)21/2=102m
Average velocity, v=ΔxΔt=102m1s=102m/s

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