Answer:
Total distance travelled by particle B for 10s
\[\tan \theta =\frac{1/2}{6}=\frac{1}{12}\]
Distance travelled by particle A for 1st 5s,
\[\theta ={{\tan }^{-1}}\left( \frac{1}{12} \right)\]
\[{{\tan }^{-1}}\left( \frac{1}{12} \right)\]
(\[XY=?\]\[XY=XZ+ZY=2XZ\text{ }or\text{ }2ZY\])
\[\Delta OZX,\] \[\sin \,60{}^\circ =\frac{XZ}{r}\]
\[=\frac{r\sqrt{3}}{2}\] Distance between\[\therefore \]
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