Question

A car travels $6km$ towards north at an angle of ${45}^o$ to the east and then travels distance of $4km$ towards north at an angle ${135}^o$ 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 ${tan}^{-1}\left(5\right)$
• B. $10km$ and ${tan}^{-1}\left(\sqrt{5}\right)$
• C. $\sqrt{52}km$ and ${tan}^{-1}\left(\sqrt{5}\right)$
• D. $\sqrt{52}km$ and ${tan}^{-1}\left(5\right)$

Answer 1

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

Net distance traveled along x-direction,

$S_X=6\cos {45}^o\hat{i}-4\cos {45}^o\hat{i}$

$=2\times \frac{1}{\sqrt{2}}=\sqrt{2}km$

Net distance traveled along y-direction

$S_y=6\sin {45}^o\hat{j}+4\sin {45}^o\hat{j}$

$=10\times \frac{1}{\sqrt{2}}=5\sqrt{2}km$

$\therefore$ Net distance traveled from the starting point,

$S=\sqrt{S_X^2+S_y^2}=\sqrt{{\left(\sqrt{2}\right)}^2+{\left(5\sqrt{2}\right)}^2}$

$=\sqrt{2+25\times 2}=\sqrt{52}$km

Angle which the resultant makes with the east direction

$\tan \theta =\frac{y}{x}=\frac{5\sqrt{2}}{\sqrt{2}}or\theta ={tan}^{-1}\left(5\right)$