Kinematics Question 22
Question: The resultant of $ \overrightarrow{P} $ and $ \overrightarrow{Q} $ perpendicular to $ \overrightarrow{P} $ . What is the angle between $ \overrightarrow{P} $ and $ \overrightarrow{Q} $
Options:
A) $ {{\cos }^{-1}}(P/Q) $
B) $ {{\cos }^{-1}}(-P/Q) $
C) $ {{\sin }^{-1}}(P/Q) $
D) $ {{\sin }^{-1}}(-P/Q) $
Correct Answer: B therefore $ \tan 90{}^\circ =\frac{Q\sin \theta }{P+Q\cos \theta } $ therefore $ P+Q\cos \theta =0 $ $ \cos \theta =\frac{-P}{Q} $ $ \theta ={{\cos }^{-1}}( \frac{-P}{Q} ) $
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