Posted by **bobby** on Thursday, January 31, 2013 at 2:29am.

heading due north at 6.00 m/s relative to the water. The local ocean current is 1.53 m/s in a direction 40° north of east. What is the velocity of the ship relative to the earth?

- physics -
**drwls**, Thursday, January 31, 2013 at 2:54am
It is the vector sum of the velocity of the ship relative to the water, and the velocity of the water with respect to the land.

- physics -
**Henry**, Friday, October 2, 2015 at 11:43pm
V = 6.0m/s[90o] + 1.53m/s[40o]

X = 6.0*Cos90 + 1.5*Cos40 = 0 + 1.15 = 1.15 m/s.

Y = 6*sin90 + 1.5*sin40 = 6 + 0.96 = 6.96 m/s.

Tan A = Y/X = 6.96/1.15 = 6.05.

A = 80.6o.

V = Y/sin A = 6.96/sin80.6 = 7.05 m/s[80.6o]

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