Question

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

• A. $\sqrt{50}km$ and ${\tan }^{-1}(5)$
• B. $10km$ and ${\tan }^{-1}(\sqrt{50})$
• C. $\sqrt{52}km$ and ${\tan }^{-1}(5)$
• D. $\sqrt{52}km$ and ${\tan }^{-1}(\sqrt{5})$

Answer 1

Answer: (C)

Net movement along x-direction
$S_x=6\cos 45^{\circ }\hat{i}+4\cos 45^{\circ }(-\overrightarrow{i})$
$S_x=(6-4)\cos 45^{\circ }\hat{i}$
$=2\times \frac{1}{\sqrt{2}}=\sqrt{2}km$

Net movement along y-direction
$S_y=6\sin 45^{\circ }\hat{j}+4\sin 45^{\circ }\hat{j}$
$S_y=(6+4)\sin 45^{\circ }\hat{j}=10\times \frac{1}{\sqrt{2}}$
$=5\sqrt{2}km$
Net movement from starting point (O in the diagram)
$|S|=\sqrt{S_x^2+S_y^2}$
$=\sqrt{(\sqrt{2}{)}^2+(5\sqrt{2}{)}^2}$
$=\sqrt{52km}$
The tan of angle which resultant makes with the east direction,
$\tan \theta =\frac{y-\text{ component }}{x-\text{ component }}$
$=\frac{5\sqrt{2}}{\sqrt{2}}$
$\Rightarrow \theta ={\tan }^{-1}(5)$