# A car is driven 80 km west and then 30 km 45 degree south of west. What is the displacement of the car from the point of origin? ( magnitude and displacement).

Mar 16, 2018

Let's break the displacement vector in to two perpendicular component i.e the vector which is $30 K m$ ${45}^{\circ}$ south of west.

So,along west component of this displacement was $30 \sin 45$ and along south this was $30 \cos 45$

So,net displacement towards west was $80 + 30 \sin 45 = 101.20 K m$

and,towards south it was $30 \cos 45 = 21.20 K m$

So,net displacement was $\sqrt{{101.20}^{2} + {21.20}^{2}} = 103.4 K m$

Making an angle of ${\tan}^{-} 1 \left(\frac{21.20}{101.20}\right) = {11.82}^{\circ}$ w.r.t west

Well this could have been solved using simple vector addition without taking perpendicular components,so I would request you to try that of your own,

Thank you :)