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A particle is moving east wards with a velocity of 5 ms^-1. In 10 sec, the velocity changes to 5 ms^-1 north wards. The average acceleration in this time is

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$\frac{1}{\sqrt{2}} {ms}^{- 2}$ towards north-east

$\frac{1}{2} {ms}^{- 2}$ towards north

Zero

$\frac{1}{\sqrt{2}} {ms}^{- 2}$ towards north-west.

answer is D.

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Detailed Solution

$\begin{array}{l} \bar{u} = 5 \bar{i}; \bar{v} = 5 \bar{j} \Rightarrow \bar{v} - \bar{u} = 5 \bar{j} - 5 \bar{i} \\ | \bar{a} | = \frac{| \bar{v} - \bar{u} |}{t} = \frac{5 \sqrt{2}}{10} = \frac{1}{\sqrt{2}} {ms}^{- 2} along N-W \end{array}$

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