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A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle is

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$\frac{1}{\sqrt{2}} m / s^{2} N - W$

$\frac{1}{\sqrt{2}} m / s^{2} S - W$

Zero

$\frac{1}{\sqrt{2}} m / s^{2} N - E$

answer is B.

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Detailed Solution

$∆ v = 2 vsin (\frac{θ}{2}) = 2 \times 5 \times \sin 45^{0} = \frac{10}{\sqrt{2}}$

$∴ a = \frac{∆ v}{∆ t} = \frac{10 \sqrt{2}}{10} = \frac{1}{\sqrt{2}} m / s^{2}$ initial direction i and final direction j change is j-i i.e. north west

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