Direction of a vector (help with part c)

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Homework Help Overview

The problem involves a model sailboat moving west with a gust of wind affecting its direction and acceleration. The context includes calculating the mass of the sailboat, its new velocity, and specifically the new direction after a gust of wind, which is described as a vector problem in physics.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss using vector addition and superposition to determine the new direction of the sailboat after the wind's influence. There are attempts to break down the components of the vectors involved and apply trigonometric functions to find angles.

Discussion Status

Some participants have provided guidance on how to approach the problem using vector components and trigonometric functions. However, there is a lack of consensus on the correct method, as some attempts have led to confusion and incorrect results. The discussion remains open with various interpretations being explored.

Contextual Notes

Participants are grappling with the correct application of vector components, particularly in relation to the angle of the wind and the direction of the sailboat. There is mention of assumptions regarding the coordinate system, specifically treating west as negative.

salaam
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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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