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5. A ship heading due west @ 1.5m/s meets a current South east at Im/s. Find the resultant speed and direction of the ship. At the same time, a crew member climbs the ship's mast at 0.7m/s, find the magnitude and direction of the Crew member travelling through space. Draw the vector diagrams to accom Σ WE Given V₁ = 1.5m/s West accompany VC = 1 m/s southeast 450 Vs N Both answers. space 273 and Component form of vectors V₁ = 1.5 m/s West A √s = -1.5 m/s. Vs I VC = 1 m/s south east Pg.4 VC = 1 m/s (cos (45)₁ - Sin (45);) vc = 10.70 -0.70, m/s →E ^ i = unit vector along X axis. j = unit veeter along y axis. kc = unit veror abng 2 axo S The resultant velocity > V = √s +V c Vs v = -1.5; +.70; -0.70; √ = -0.8; -0.70j Iresultant speed V = √0.82 +0.702 V= V = √0.64 +0.49 V = √113 V = 1.063 m/s Direction 0.70 73:00 0 =arcton (0.80 • 41.19° South of west.

5. A ship heading due west @ 1.5m/s meets a current South east at Im/s. Find the resultant speed and direction of the ship. At the same time, a crew member climbs the ship's mast at 0.7m/s, find the magnitude and direction of the Crew member travelling through space. Draw the vector diagrams to accom Σ WE Given V₁ = 1.5m/s West accompany VC = 1 m/s southeast 450 Vs N Both answers. space 273 and Component form of vectors V₁ = 1.5 m/s West A √s = -1.5 m/s. Vs I VC = 1 m/s south east Pg.4 VC = 1 m/s (cos (45)₁ - Sin (45);) vc = 10.70 -0.70, m/s →E ^ i = unit vector along X axis. j = unit veeter along y axis. kc = unit veror abng 2 axo S The resultant velocity > V = √s +V c Vs v = -1.5; +.70; -0.70; √ = -0.8; -0.70j Iresultant speed V = √0.82 +0.702 V= V = √0.64 +0.49 V = √113 V = 1.063 m/s Direction 0.70 73:00 0 =arcton (0.80 • 41.19° South of west.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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A ship heading due west @ 1.5 m/s meets a current going South east at 1m/s. Find the resultant speed and direction of the ship. At the same time, a crew member climbs Heship's mast at 0.7m/s, find the magnitude and direction of the crew member travelling through space: draw the space and vector diagrams to accompany both answers.  

 

not sure if this example I have is acceptable or if the answer is right. I'm struggling to understand all the vectors and units. So better space and vector diagrams and a break down of the formulas and units would be great thanks 

5. A ship heading due west @ 1.5m/s meets a current South east at Im/s. Find the resultant speed and direction of the ship. At the same time, a crew member climbs the ship's mast at 0.7m/s, find the magnitude and direction of the Crew member travelling through space. Draw the vector diagrams to accom Σ WE Given V₁ = 1.5m/s West accompany VC = 1 m/s southeast 450 Vs N Both answers. space 273 and Component form of vectors V₁ = 1.5 m/s West A √s = -1.5 m/s. Vs I VC = 1 m/s south east Pg.4 VC = 1 m/s (cos (45)₁ - Sin (45);) vc = 10.70 -0.70, m/s →E ^ i = unit vector along X axis. j = unit veeter along y axis. kc = unit veror abng 2 axo S The resultant velocity > V = √s +V c Vs v = -1.5; +.70; -0.70; √ = -0.8; -0.70j Iresultant speed V = √0.82 +0.702 V= V = √0.64 +0.49 V = √113 V = 1.063 m/s Direction 0.70 73:00 0 =arcton (0.80 • 41.19° South of west.
Transcribed Image Text:5. A ship heading due west @ 1.5m/s meets a current South east at Im/s. Find the resultant speed and direction of the ship. At the same time, a crew member climbs the ship's mast at 0.7m/s, find the magnitude and direction of the Crew member travelling through space. Draw the vector diagrams to accom Σ WE Given V₁ = 1.5m/s West accompany VC = 1 m/s southeast 450 Vs N Both answers. space 273 and Component form of vectors V₁ = 1.5 m/s West A √s = -1.5 m/s. Vs I VC = 1 m/s south east Pg.4 VC = 1 m/s (cos (45)₁ - Sin (45);) vc = 10.70 -0.70, m/s →E ^ i = unit vector along X axis. j = unit veeter along y axis. kc = unit veror abng 2 axo S The resultant velocity > V = √s +V c Vs v = -1.5; +.70; -0.70; √ = -0.8; -0.70j Iresultant speed V = √0.82 +0.702 V= V = √0.64 +0.49 V = √113 V = 1.063 m/s Direction 0.70 73:00 0 =arcton (0.80 • 41.19° South of west.
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