Question

A car travels $6\text{km}$ towards north at an angle of ${45}^{\circ }$ to the east and then travels distance of $4\text{km}$ 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. $\sqrt{50}\text{km}$ and ${\tan }^{-1}\left(5\right)$
• B. $10\text{km}$ and ${\tan }^{-1}\left(\sqrt{5}\right)$
• C. $\sqrt{52}\text{km}$ and ${\tan }^{-1}\left(5\right)$
• D. $\sqrt{52}\text{km}$ and ${\tan }^{-1}\left(\sqrt{5}\right)$

Answer 1

The correct option is C $\sqrt{52}\text{km}$ and ${\tan }^{-1}\left(5\right)$

Net distance travellel along x-direction,
$S_x=6\cos {45}^{\circ }\hat{i}-4\cos {45}^{\circ }\hat{i}$
$=2\times \frac{1}{\sqrt{2}}=\sqrt{2}\text{km}$

Net distance travelled along y - direction
$S_y=6\sin {45}^{\circ }\hat{j}+4\sin {45}^{\circ }\hat{j}$
$=10\times \frac{1}{\sqrt{2}}=5\sqrt{2}\text{km}$
$\therefore$ Net distance travelled from the starting point,
$S=\sqrt{S_x^2+S_y^2}=\sqrt{(\sqrt{2}{)}^2+(5\sqrt{2}{)}^2}$
$=\sqrt{2+25\times 2}=\sqrt{52}\text{km}$
Angle which the resultant makes with the east direction
$\tan \theta =\frac{y}{x}=\frac{5\sqrt{2}}{\sqrt{2}}$ or $\theta ={\tan }^{-1}\left(5\right)$