Question

A car travels 6 km towards north at an angle of 45° to the east and then travels distance of 4 km towards north at an angle of 135° to the 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. 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)

$\sqrt{52}km$and$\tan^{- 1}(5)$

Answer 2

Net movement along x-direction S_x = (6 - 4) cos 45° $\widehat{i}$ $= 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}km$

Net movement along y-direction S_y = (6 + 4) sin 45°$\widehat{j}$ $= 10 \times \frac{1}{\sqrt{2}} = 5\sqrt{2}km$

Net movement from starting point $| \vec { s } | = \sqrt { s _ { x } ^ { 2 } + s _ { y } ^ { 2 } }$

$= \sqrt{\left( \sqrt{2} \right)^{2} + \left( 5\sqrt{2} \right)^{2}}$=$\sqrt{52}km$

Angle which makes with the east direction

$\tan\theta = \frac{Y - \text{component}}{X - \text{component}}$ $= \frac{5\sqrt{2}}{\sqrt{2}}$

\f0 $\theta = \tan^{- 1}(5)$