Calculating Velocity and Direction of a Sailboat After a Sudden Gust of Wind

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Homework Help Overview

The discussion revolves around calculating the velocity and direction of a sailboat after experiencing a sudden gust of wind. The problem involves concepts from kinematics, specifically focusing on acceleration, velocity, and vector resolution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore how to calculate the final velocity and direction of the sailboat after the gust subsides, questioning whether to consider the initial velocity as zero or 5 m/s. There are discussions about vectorizing the change in velocity and resolving acceleration into x and y components.

Discussion Status

Some participants have offered guidance on resolving the acceleration into components and applying kinematic equations, while others express confusion about the setup and calculations. There is a mix of interpretations regarding the application of time to the different questions posed.

Contextual Notes

Participants note the assumption of north as the positive y-direction and east as the positive x-direction. There is uncertainty about the initial conditions and how to visualize the problem effectively.

viendong
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A sailboat is traveling east at 5. m/s. A sudden gust of wind gives the boat an acceleration =0.80 m/s^2, 40 degree north of east
1) what is the boat's speed later when the gust subsides?
2) what is the boat's direction 6.0 s later when the gust subsides?

I've a question
I need to find the velocity when the gust wind
so
[text]vf = vi +at [\text]
do the vi =0 or it's 5 m/s though?
how should i visualize the problem?
Thanks again
 
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I'm not really able to see how you can solve it if the 6.0s does not also apply to question 1)
If so then you can calculate the change in velocity.
You can then vectorize this change and see how it effects the x and y velocity.
Since you have 5 m/s as x you can see how they affect each other and you will get a new x vector. Combine the new x vector with the old y vector to get the final vector which has the boat's velocity and direction.
Maybe there's an easier way to do it? Not sure though, are you working with vectors now?
 
Assume the north to be the positive y-direction and east to be the positive x. Resolve the acceleration in both the directions and then apply the equation

[tex]v = u +at[/tex] for each direction.

So, for north, you have the inital velocity as zero while it is 5m/s for east. After calculating the final velocities in each direction, use pythagoras theorem and find the final velocity.

The direction is given by [tex]\theta= tan^{-1}\frac{vy}{vx}[/tex]

[tex]\Theta[/tex] is the angle made with the east.

spacetime
www.geocities.com/physics_all/index.html
 
i still don't get how can I resolve to find the acceleration for each of (x, y) direction
Isn't x acceleration =0 and the y direction is .80 m/s^2 ? I'm confused... please help
 
thank you very much, i got it now .. hihii :)
 

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