Question

A car travels $6\, km$ towards north at an angle of $45^{\circ}$ to the east and then travels distance of $4 \, km$ towards north at an angle $135^{\circ}$ to east. 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?

• A. $\sqrt{50} \, km$ and $\tan ^{-1}(5)$
• B. $10 \, km$ and $\tan ^{-1}(\sqrt{5})$
• 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 =(6-4) \cos 45^{\circ} \hat{i} $
$ =2 \times \frac{1}{\sqrt{2}}=\sqrt{2} km$
Net movement along $y$-direction
$S_y =(6+4) \sin 45^{\circ} \hat{j}$
$ =10 \times \frac{1}{\sqrt{2}}=5 \sqrt{2} km$
Net movement from starting point
$|\vec{S}| =\sqrt{S_x^2+S_y^2}=\sqrt{(\sqrt{2})^2+(5 \sqrt{2})^2}$
$ =\sqrt{52} km$
Angle which resultant makes with the east direction
$\tan \theta =\frac{y \text {-component }}{x \text {-component }}$
$=\frac{5 \sqrt{2}}{\sqrt{2}}$
$\therefore \theta =\tan ^{-1}(5)$