A particle moves towards east with velocity 5 m/s . After 10 seconds its direction changes towards n

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3-2.Motion in Plane

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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

A

Zero

B

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

C

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

D

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

Solution

$\Delta v=2 v \sin \left(\frac{\theta}{2}\right)$

$=2 \times 5 \times \sin 45^{\circ}=\frac{10}{\sqrt{2}}$

$\therefore a=\frac{\Delta v}{\Delta t}=\frac{10 / \sqrt{2}}{10}=\frac{1}{\sqrt{2}} \mathrm{m} / \mathrm{s}^{2}$

Standard 11

Physics

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