Answer to Question #23278 in Mechanics | Relativity for Syomara

Question

a ship leaves port at 1 pm traveling north at the speed of 30 miles per hour. At 3 pm the ship adjust its course 20 degrees eastward, how far was the ship from the port at 4pm.

Answer 1

The displacement from 1 pm to 3pm
D1 = 30 miles/hr * 2 hr = 60 miles
to the North

The displacement from 3 pm till 4 pm
D2 = 30 miles/hr * 1 hr = 30 miles
to the 20 degrees North Eastward

The resulting displacement, by adding two vectors
D = sqrt((D1 + D2*cos(20 deg))^2 + (D2*sin(20 deg))^2)
D = sqrt((60 + 30*cos(20 deg))^2 + (30*sin(20 deg))^2) = 88.79 miles

Answer: about 88.8 miles away from the port.