Question

A car travels $6km$ towards north at an angle of $45^{\circ }$ to the east and then travels distance of $4km$ 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. $10km$ 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
$\tan \theta =\frac{y\text{-component }}{x\text{-component }}$
$=\frac{5\sqrt{2}}{\sqrt{2}}$
$\therefore \theta ={\tan }^{-1}(5)$