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A particle is moving towards east with velocity 5 m/s. In 10 seconds its direction changes towards north speed remaining unchanged. The average acceleration of the particle is

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a

Zero

b

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

c

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

d

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

answer is B.

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Detailed Solution

$\vec{v_{i}} = v \hat{i} a n d \overset{⇀}{v_{f}} = v \hat{j} T h e r e f o r e ∆ \vec{v} = \vec{v_{f}} - \vec{v_{i}} = (v \hat{j} - v \hat{i}) \Rightarrow ∆ v = \sqrt{2} v \Rightarrow a = \frac{\sqrt{2} \times 5}{10} m / s^{2} = \frac{1}{\sqrt{2}} m / s^{2}, n o r t h w e s t .$

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