How does wind acceleration affect the speed and direction of a sailboat?

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Homework Help Overview

The discussion revolves around a sailboat's motion affected by wind acceleration. The original poster presents a scenario where a sailboat is initially traveling east at a specific speed and experiences a gust of wind that provides acceleration at an angle. The goal is to determine the boat's speed and direction after a set time period.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial conditions of the problem, including the boat's speed and the acceleration due to wind. There are attempts to break down the acceleration into its components and calculate the resulting velocity after a certain time. Some participants express confusion about the relationship between velocity and acceleration.

Discussion Status

There is an ongoing examination of the calculations presented by the original poster, with some participants pointing out potential errors in mixing concepts of velocity and acceleration. The discussion includes suggestions to clarify the effective acceleration and how to apply it to find the final velocity components. Multiple interpretations of the problem are being explored.

Contextual Notes

Some participants note that there may be insufficient information to solve the problem as stated, and there is a mention of a possible trick question regarding the initial speed remaining unchanged. The need for clarity on the time of acceleration is also highlighted.

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Homework Statement


A sailboat is traveling east at 4.5ms-1. a suddent gust of wind gives the boat an acceleration a = 0.70 ms-2, 40 degress north to east. what are the boat's speed and direction 6s later when the gust subsides?

Homework Equations


V(@6s) = V(init.) + at

The Attempt at a Solution


i seriously have not a clue where to start. tried to look it up in books and online but i still don't know what to do with the acceleration.
so the a(x) = 0.70cos40 V(x) = 4.5
a(y) = 0.70sin40 V(y) = 0 am i right?​
thus, V(6s)(x) = 4.5 + .070cos40(6) = 7.7174
V(6s)(y) = 0 + 0.70sin40(6) = 3.8567​
therefore, direction n in degree:
tan(n) = 3.8567 / 7.7174, n = 26.55 degrees​
velocity = (7.7174^2 + 3.8567^2)^(1/2) = 8.6274​
and i am told that i am wrong. hope you guys can point out why. thanks!
 
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It does not appear to me that you have enough inforation to solve the problem. You need the time of the acceleration.

Unless it is a trick question and the answer is 4.5m/s, East...

Anyway, your analysis is certainly wrong as you are mixing velocity and acceleration.
 
oops sorry i left that out. just edited the thread.
 
Well, then the answer is that you mixed acceleration and speed. You need to use the acceleration and the time to calculate the magnitude of the second speed vector before adding the two vectors together.
 
ere said:
so the a(x) = 0.70cos40 V(x) = 4.5
a(y) = 0.70sin40 V(y) = 0 am i right?​
thus, V(6s)(x) = 4.5 + .070cos40(6) = 7.7174
V(6s)(y) = 0 + 0.70sin40(6) = 3.8567​
therefore, direction n in degree:
tan(n) = 3.8567 / 7.7174, n = 26.55 degrees​
velocity = (7.7174^2 + 3.8567^2)^(1/2) = 8.6274​
and i am told that i am wrong. hope you guys can point out why. thanks!

What is wrong? and Where? What do you mean by confusing velocity and acceleration? Find the effective acceleration for both Y and X axis, then use the
linear laws of motion to get the final velocity components? Why is this wrong?
 
ere said:
V(6s)(y) = 0 + 0.70sin40(6) = 3.8567[/indent]

This calculation seems to have error
 

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