question_answer

A person walks at the rate of \[3\,km/hr\]. Rain appears to him in vertical direction at the rate of \[3\sqrt{3}km/hr\]. Find the magnitude and direction of true velocity of rain.

A)  \[6\,km/hr\], inclined at an angle of \[45{}^\circ \] to the vertical towards the person's motion.

B)  \[3\,km/hr\], inclined at an angle of \[30{}^\circ \] to the vertical towards the person's motion.

C)  \[6\,km/hr\], inclined at an angle of \[30{}^\circ \] to the vertical towards the person's motion.

D)  \[6\,km/hr\], inclined at an angle of \[60{}^\circ \] to the vertical towards the person's motion.

Correct Answer: C

Solution :

[c]  \[{{\vec{v}}_{r/m}}={{\vec{v}}_{r}}-{{\vec{v}}_{m}}\] \[{{\vec{v}}_{r}}={{\vec{v}}_{r/m}}+{{\vec{v}}_{m}}\] \[\left| {{{\vec{v}}}_{r}} \right|=\sqrt{v_{r/m}^{2}+v_{m}^{2}}\] \[=\sqrt{{{(3)}^{2}}+{{(3\sqrt{3})}^{2}}}=6km/hr\] \[\tan \theta =\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}\Rightarrow \theta =30{}^\circ \]