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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20 m/s

$(10 / \sqrt{2}) m / s$

5 m/s

$10 \sqrt{2} m / s$

answer is C.

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Detailed Solution

The net displacement will be
$\begin{array}{l} Δx = (10 \hat{j} - 10 \hat{i}) m \\ | Δx | = {[(10)^{2} + (10)^{2}]}^{1 / 2} = 10 \sqrt{2} m \end{array}$
Average velocity, $v = \frac{Δx}{Δt} = \frac{10 \cdot \sqrt{2} m}{1 s} = 10 \sqrt{2} m / s$

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