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A 358-kg boat is sailing 15.9° north of east at a speed of 1.80 m/s. Thirty seconds later, it is sailing 36.6° north of east at...Asked by Natasha
                A 343-kg boat is sailing 16.2° north of east at a speed of 1.98 m/s. Thirty seconds later, it is sailing 37.5° north of east at a speed of 3.76 m/s. During this time, three forces act on the boat: a 30.3-N force directed 16.2° north of east (due to an auxiliary engine), a 22.3-N force directed 16.2° south of west (resistance due to the water), and W (due to the wind). Find the magnitude and direction of the force W. Express the direction as an angle with respect to due east.
            
            
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                    Answered by
            drwls
            
    Compute the velocity change vector during the 30 seconds and divide it by 30 for the average acceleration vector. 
Multiply that by the mass for the net force vector, Fnet. You know two of the three vectors that make up Fnet. Use that to determine the third (Wind) force vector
    
Multiply that by the mass for the net force vector, Fnet. You know two of the three vectors that make up Fnet. Use that to determine the third (Wind) force vector
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