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. Sketc
1 answer:
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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