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

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The discussion focuses on calculating the speed and direction of a sailboat after experiencing a gust of wind that accelerates it at 0.70 m/s² at an angle of 40 degrees north of east. The initial speed is 4.5 m/s east, and the calculations involve breaking down the acceleration into x and y components. Participants point out that the original calculations mix velocity and acceleration, leading to incorrect results. The correct approach requires determining the effective acceleration for both axes and applying linear motion equations to find the final velocity. Overall, the thread emphasizes the importance of accurately distinguishing between acceleration and velocity in physics problems.
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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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