Asked by Joe
Matt is attempting to sail a sailboat northeast across the river with a velocity of 13 m/s relative to the
current of water. The water current is flowing from south to north at a velocity of 6 m/s. If the angle between
the velocity of the sailboat and the water current is 40 degrees
, what is the resultant or true velocity of the sailboat?
current of water. The water current is flowing from south to north at a velocity of 6 m/s. If the angle between
the velocity of the sailboat and the water current is 40 degrees
, what is the resultant or true velocity of the sailboat?
Answers
Answered by
Scott
N component ... 6 m/s + 13cos(40º) m/s
E component ... 13sin(40º) m/s
(resultant)^2 = N^2 + E^2
E component ... 13sin(40º) m/s
(resultant)^2 = N^2 + E^2
Answered by
Scott
N component ... 6 m/s + 13cos(40º) m/s
E component ... 13sin(40º) m/s
(resultant)^2 = N^2 + E^2
tan(direction) = N / E
E component ... 13sin(40º) m/s
(resultant)^2 = N^2 + E^2
tan(direction) = N / E
Answered by
Mark
Hey thanks, but could you explain it a little? But also would the final resultant be 13sin(40)m/s? Why would we not use cos in this instant?
Answered by
Mark
Actually I think I got it. So I could just add both N component and E component together to get a final resultant of 22.911 N of E then correct? Thanks
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