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 positions makes with the cast?

• 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
$S_x=(6-4)cos45^{\circ }\hat{i}$
$=2\times \frac{1}{\sqrt{2}}=\sqrt{2}km$
Net movement along Y -direction
$S_y=(6+4)sin45^{\circ }\hat{j}$
$=10\times \frac{1}{\sqrt{2}}=5\sqrt{2}km$
Net movement from starting point
$|S|=\sqrt{S_x^2+S_y^2}=\sqrt{(\sqrt{2}{)}^2+(5\sqrt{2}{)}^2}$
$=\sqrt{52}km$
Angle which resultant makes with the east direction
$tan\theta =\frac{y-component}{x-component}$
$=\frac{5\sqrt{2}}{\sqrt{2}}$
$\therefore 40mm\theta ={\tan }^{-1}(5)$