Direction of a vector (help with part c)

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The discussion revolves around calculating the new direction of a sailboat after experiencing a gust of wind. The initial velocity of the sailboat is 0.33 m/s west, and the wind provides a constant acceleration of 0.30 m/s² at an angle of 28 degrees south of west. Participants suggest using vector superposition to combine the sailboat's velocity and the wind's acceleration, breaking them into their x and y components. Despite attempts to apply trigonometric functions and arctangent calculations, the user struggles to arrive at the correct angle for the new direction. The conversation emphasizes the importance of accurately breaking down vectors and suggests revisiting the calculations for clarity.
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Homework Statement


A model sailboat is slowly sailing west across a pond at .33 m/s a gust of wind blowing at 28 degrees south of west gives the sailboat a constant acceleration of magnitude .30 m/s^2 during a time interval of 2.0 s (a) if the net force on the sailboat during the 2.0 s interval has magnitude .375 N, what is the sailboat's mass? (b) what is the new velocity of the boat after the 2.0-s gust of wind? (c) what is the new direction of the boat after the 2s gust of wind... ___ south of west.


Homework Equations


f=ma
vf=vo+at
tan^-1=vy/vx


The Attempt at a Solution


i have already determined a using f=ma .375 / .3 which is 1.25 kg

i have part b... by using vf= vo + at which is .930 m/s

but I am very confused with c! i think that i should be using tan^-1 (Vy/Vx) but i have no idea what to plug in.
 
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Use superpostion. And ad the vectors up. One vector has a magnutiude of .33m/s @ 0 degrees. The second is the wind blowing on the boat then use your arctan to find the angle
 
what do you mean superposition? well i understand that cos(0)(.33) is .33, sin(0)(.33) is 0. and then i don't understand what to do with the 28 degrees
 
You are on the right path. The cos(0)*.33=.33 that gives you the x component of V1. Sin(0)*.33 gives you the y component of V1. Now take the magnutiude of V2 (.3) and break it down into its X & Y component using -28 degrees (its negative because it south aka below the x axis). The add the X component of V1 and V2 together and do the same for the Y component and uses those vaules to find the new direction.
 
that makes sense.. and i tried what you said but it is incorrect! any other ideas?
i added .33 with .264 and 0 with -.1408 i took the arctan of (.594/-.1408) and got -76.66
 
try negative.33 and negative .3 since I would consider west negative x
 
that did not work either :( please help!
 

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