A) \[5\sqrt{5}\,km/h,\,ta{{n}^{-1}}(1/2)\,S\,of\,E\]
B) \[5\sqrt{5}\,km/h,\,ta{{n}^{-1}}(1/2)\,E\,of\,S\]
C) \[4\sqrt{5}\,km/h,\,ta{{n}^{-1}}(1/2)\,S\,of\,E\]
D) \[4\sqrt{5}\,km/h,\,ta{{n}^{-1}}(1/2)\,E\,of\,S\]
Correct Answer: A
Solution :
[a] Here we are given velocity of 'A', \[{{\vec{v}}_{A}}=10\hat{i}\] Velocity of 'A', w.r.t. 'B', \[{{\vec{v}}_{A/B}}=5j\] Now \[{{\vec{v}}_{A/B}}={{\vec{v}}_{A}}-{{\vec{v}}_{B}}\] \[5\hat{j}=10\hat{i}-{{\vec{v}}_{B}}\Rightarrow {{\vec{v}}_{B}}=10\hat{i}-5\hat{j}\] Hence velocity of B, \[{{v}_{B}}=\sqrt{{{10}^{2}}+{{5}^{2}}}=5\sqrt{5}km/h\] \[\tan \theta =\frac{5}{10}=\frac{1}{2}\] \[\theta ={{\tan }^{-1}}\left( \frac{1}{2} \right)s\]of E
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