Question

A car travels $6\, km$ towards the north at an angle $45^{\circ}$ to the east and then travels distance of $4\, km$ 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. $10 km$ 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
$S_{y}=6 \sin 45^{\circ} \hat{j}+4 \sin 45^{\circ} \hat{j}$
$S_{y}=(6+4) \sin 45^{\circ} \hat{j}=10 \times \frac{1}{\sqrt{2}}$
$=5 \sqrt{2} k m$
Net movement from starting point (O in the diagram)
$|S|=\sqrt{S_{x}^{2}+S_{y}^{2}}$
$=\sqrt{(\sqrt{2})^{2}+(5 \sqrt{2})^{2}}$
$=\sqrt{52 k m}$
The tan of angle which resultant makes with the east direction,
$\tan \theta =\frac{y-\text { component }}{x-\text { component }}$
$=\frac{5 \sqrt{2}}{\sqrt{2}}$
$\Rightarrow \theta=\tan ^{-1}(5)$