Kinematics Question 113

Question: A particle moves towards east with velocity 5 m/s. After 10 seconds is direction changes towards north with same velocity. The average acceleration of the particle is [CPMT 1997; IIT-JEE 1982]

Options:

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 $

Show Answer

Answer:

Correct Answer: B

Solution:

$ \Delta v=2v\sin ( \frac{\theta }{2} )=2\times 5\times \sin 45{}^\circ $ = $ \frac{10}{\sqrt{2}} $ \ $ a=\frac{\Delta v}{\Delta t}=\frac{10/\sqrt{2}}{10}=\frac{1}{\sqrt{2}}\ m/s^{2} $



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