Question

A car travels $6km$ towards north at an angle of ${45}^o$ to the east and then travels distance of $4km$ towards north at an angle of ${135}^o$ to the east (Figure). How far is the point from the starting point? What angle does the straight: line joining its initial and final position makes with the east?

Answer 1

Net movement along $x$-direction,

$S_x=(6-4)\cos {45}^o\hat{i}=2\times \frac{1}{\sqrt{2}}=\sqrt{2}km$

Net movement along $y$-direction,

$S_y=(6+4)\sin {45}^o\hat{j}=10\times \frac{1}{\sqrt{2}}=5\sqrt{2}km$

Net movement from startinf point,

$\overrightarrow{s}|=\sqrt{S_x^2+S_y^2}=\sqrt{(\sqrt{2}{)}^2+(5\sqrt{2}{)}^2}=\sqrt{52}km$

Angle which makes with the east direction,

$\tan \theta =\frac{\text{Y-component}}{\text{X-component}}=\frac{5\sqrt{2}}{\sqrt{2}}$

$\therefore \theta ={\tan }^{-1}\left(5\right)$