# Sailboat Direction - Kinematics in Two Directions

• bimbambaby
In summary, the sailboat is initially traveling east at 5.1m/s and experiences a gust of wind with an acceleration of 0.80 m/s2 at 40° north of east. After 6 seconds, the boat's direction is calculated to be 19° north of east. However, this answer may be incorrect due to potential misinterpretation of the angle of reference. The correct answer is likely simply east, as the boat has stopped once the gust subsides.

## Homework Statement

A sailboat is traveling east at 5.1m/s . A sudden gust of wind gives the boat an acceleration = (0.80 m/s2}, 40° north of east).

What is the boat's direction 6s later when the gust subsides?

Vf = V0 + at

## The Attempt at a Solution

To solve I simply created X and Y components of the velocity after those six seconds, in order to use the arctan function to then find my angle:

Vfx = Vox+at
Vfx = 5.1 m/s + (6s)* (0.80cos40) = 8.777013327 m/s (x)

Vfy = 0 m/s +(6s) * (0.80sin40) = 3.085380526 m/s (y)

To find the angle, I then simply input these values into the arctan function:

tan-1(vy/vx) = θ
tan-1(3.085/8.777) = 19°, north of east

This answer, however, is incorrect. Do you have any advice on my approach, or anything that I may be negating my calculations?

Ok here is just a suggestion. Try using 50 as the angle and recalculate the velocities. This I think they were trying to say that the angle of the wind gust it makes with the north is 40 degrees.

I appreicate that, but there is a first part of the question that asks for the velocity after 7.7 seconds, in which I used 40° as my angle of reference. I arrived at the correct answer for that. It also says "north of east," which I am sure means north of the negative x-axis (east). I believe that were it phrased otherwise (east of north), that would mean 40° east (to the left) of the positive y-axis, so I don't believe the angles are the issue, unless someone else feels otherwise.

No, "40 degrees N of east" means starting from the east (the x-axis) measure toward the north (counter clockwise) 40 degrees. That is what bimbambaby is doing.

Halls of Ivy -

Any idea why the method I employed wouldn't work? I've calculated my velocity components correctly, and that should provide the correct answer, in theory.

The situation is not like of projectile or free fall.
Once the engine and the wind stops, we assume it to be stopped(maybe it takes a few seconds)

I think the question ask is "where", in direction of x and y.

Azizlwl -

I'm not quite sure that I follow your meaning? We're determining the direction of the sailboat in terms of angle relative to the negative (east) x-axis. That is the directional component of the vector.

We assume the engine is giving the boat a constant velocity due east.
The wind too giving a force that equivalent to acceleration.

We can find the position of the boat after 6s.

After the gust subsides surely the boat is heading east.