Answer to Question #191406 in Physics for Shahir

Question

8) An object moved 200 m E, 50 m 30° N of W , then 20 m 45 ^ 0 of W. Find the final displacement

Answer 1

$$s=(200,0)+(-50\cos{30},50\sin{30})\\+(-20\cos{45},20\sin{45})=(142.6,39.1)\ m\\s=\sqrt{142.6^2+39.1^2}=148\ m\\ \theta=\arctan{\frac{39.1}{142.6}}=15\degree \text{ N of E}$$