Asked by Bob
a ship leaves portland sails northwest for 1 hour and then northeast for 2 hours. If it does not change speed, find what course the ship should take to return directly to port. also find how long the return will take
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Let the ship's speed be v. Put the origin at the port, take +x = east, +y = north.
First leg (1 h) northwest: displacement
d1 = v(−√2/2, √2/2).
Second leg (2 h) northeast: displacement
d2 = v·2(√2/2, √2/2) = v(√2, √2).
Total displacement from port:
R = d1 + d2 = v(√2/2, 3√2/2) = v(√2/2)(1,3).
To return directly to port the ship must travel −R = v(√2/2)(−1,−3). The direction of −R is given by the ratio of components: |x|:|y| = 1:3. Thus the heading is 18.4349° west of south (equivalently S 18°26'6" W), i.e. bearing 198.435°.
Distance to port = |R| = v√5, so time to return = distance/speed = √5 hours ≈ 2.236 h.
Answer: steer S 18.435° W (bearing 198.435°) and it will take √5 ≈ 2.236 hours.
First leg (1 h) northwest: displacement
d1 = v(−√2/2, √2/2).
Second leg (2 h) northeast: displacement
d2 = v·2(√2/2, √2/2) = v(√2, √2).
Total displacement from port:
R = d1 + d2 = v(√2/2, 3√2/2) = v(√2/2)(1,3).
To return directly to port the ship must travel −R = v(√2/2)(−1,−3). The direction of −R is given by the ratio of components: |x|:|y| = 1:3. Thus the heading is 18.4349° west of south (equivalently S 18°26'6" W), i.e. bearing 198.435°.
Distance to port = |R| = v√5, so time to return = distance/speed = √5 hours ≈ 2.236 h.
Answer: steer S 18.435° W (bearing 198.435°) and it will take √5 ≈ 2.236 hours.
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