A car going due North at $20 \sqrt{2} m s^{- 1}$ turns right through an angle of $90^{0}$ without changing speed. The change in velocity of car is

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$20 m s^{- 1}$ in South East direction

$20 \sqrt{2} m s^{- 1}$ in South – East direction

$20 m s^{- 1}$ in North-East direction

$40 m s^{- 1}$ in South-East direction

answer is A.

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Detailed Solution

Change in velocity is

$Δ \bar{V} = {\bar{V}}_{f} - {\bar{V}}_{i} = 10 \sqrt{2} \hat{i} - 10 \sqrt{2} \hat{j}$

$|Δ V| = 20 m s^{- 1}$

and direction is south - east

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