Kinematics Question 521

Question: A person walks at the rate of $ 3km/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.

Options:

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

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

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

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

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Answer:

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}} $

$ | {{{\vec{v}}} _{r}} |=\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 $



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