Question

A ship leaves port at 1 pm traveling north at the speed of 30 miles/hour .at 3 pm the ship adjusts its course20° eastward. Sketch a diagram to show the path the ship moved and the distance the ship is from the port. And how far the ship is from the port at 4pm?

Answer 1

The ship is 88.8 miles far from the port at 4 pm.

Given,

The displacement from 1 to 3 in the afternoon.

D1 = 30 miles/hr × 2 hr

     = 60 miles to the north

The displacement from 3 till 4 in the afternoon.The ship changes its course 20 degrees eastward at 3 o'clock.

Therefore, D2 = 30 miles/hr × 1 hr

                       = 30 miles to the 20° north eastward

By combining two vectors, the resulting displacement:

D = √((D1 ₊ D2 × cos(20°))² ₊ (D2 × sin(20°))²

D = √((60 ₊ 30 × cos (20°))² ₊ (30 × sin(20°))²

D = 88.8 miles

Hence the ship is 88.8 miles far away from the port at 4 pm.

Learn more about "Law of sines and cosines" here-

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