Question

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

• A. $\sqrt{50}\text{km}$ and $ta{n}^{-1}\left(5\right)$
• B. $\sqrt{10}\text{km}$ and $ta{n}^{-1}\left(\sqrt{5}\right)$
• C. $\sqrt{52}\text{km}$ and $ta{n}^{-1}\left(5\right)$
• D. $\sqrt{52}\text{km}$ and $ta{n}^{-1}\left(\sqrt{5}\right)$

Answer 1

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

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