question_answer

A car travels 6 km towards north at an angle of 45° to the east and then travels distance of 4 km towards north at an angle 135° to east. How far is the point from the starring point? What angle does the straight line joining its initial and final position makes with the east?

A)  \[\Omega \] km and \[{{\tan }^{-1}}(5)\]

B)  10 km and \[{{\tan }^{-1}}\left( \sqrt{5} \right)\]

C)  \[\sqrt{50}\] km and \[{{\tan }^{-1}}(5)\]

D)  \[\sqrt{5}\]km and \[{{\tan }^{-1}}\left( \sqrt{5} \right)\]

Correct Answer: C

Solution :

Net movement along x-direction \[\Rightarrow \] \[F=ma\] Net movement along y-direction \[=m\frac{{{d}^{2}}x}{d{{t}^{2}}}\] \[=-m{{\omega }^{2}}x\] Net movement from starting point \[\omega =\sqrt{\frac{3kx}{m}}\] \[\Rightarrow \] Angle which resultant makes with the east   direction \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{3kx}}\] \[=2\pi \sqrt{\frac{m}{3k(sin\omega t)}}\] \[\Rightarrow \]  \[T\propto \frac{1}{\sqrt{a}}\]