Vectors Question 83
Question: 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
[CPMT 1997; IIT-JEE 1982]
Options:
A) Zero
$ \frac{1}{\sqrt{2}},m/s^{2},N$
C) $ \frac{1}{\sqrt{2}},m/s^{2},N-E $
D) $ \frac{1}{\sqrt{2}},m/s^{2},S-W $
Correct Answer: B $ \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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Answer:
Solution:
BETA
