Question

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).

Answer 1

Let's break the displacement vector in to two perpendicular component i.e the vector which is `30Km` `45^@` 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=101.20Km`

and,towards south it was `30 cos 45=21.20Km`

So,net displacement was `sqrt(101.20^2 +21.20^2)=103.4 Km`

Making an angle of `tan^-1(21.20/101.20)=11.82^@` 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 :)