# Answer to Question #23278 in Mechanics | Relativity for Syomara

Question #23278

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.

Expert's answer

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.

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.

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