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A particle moving with a velocity equal to 0.4 ms^-1 is subjected to an acceleration of 0.15 ms^-2 for 2 seconds in a direction at right angles to the direction of motion. The magnitude of the final velocity is

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0.3 ms^-1

0.7 ms^-1

0.5 ms^-1

0.4 ms^-1

answer is C.

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Detailed Solution

$\begin{array}{l} \bar{v} = \bar{u} + \bar{a} t = 0.4 \hat{i} + (0.15 \hat{j}) (2) = 0.4 \hat{i} + 0.3 \hat{j} \\ v = \sqrt{(0.4)^{2} + (0.3)^{2}} = 0.5 {ms}^{- 1} \end{array}$

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