Direction of a vector (help with part c)

In summary, the sailboat experiences a constant acceleration of magnitude .30 m/s^2 during a time interval of 2.0 s. The sailboat's mass is 1.25 kg after the gust of wind. The boat's new velocity is .930 m/s after the gust of wind and the boat's new direction is south of west.
  • #1
salaam
18
0

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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  • #2
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
 
  • #3
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
 
  • #4
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.
 
  • #6
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
 
  • #7
try negative.33 and negative .3 since I would consider west negative x
 
  • #8
that did not work either :( please help!
 

1. What is the direction of a vector?

The direction of a vector is the angle at which the vector is pointing in relation to a reference point or axis.

2. How is the direction of a vector represented?

The direction of a vector can be represented using either a unit vector, which has a magnitude of 1 and points in the same direction as the original vector, or using an angle measured counterclockwise from a reference axis.

3. How do you find the direction of a vector given its components?

To find the direction of a vector given its components, you can use the inverse tangent function (arctan) to calculate the angle between the vector and the x-axis. The formula is θ = arctan(y/x), where y is the vertical component and x is the horizontal component of the vector.

4. Can a vector have a negative direction?

Yes, a vector can have a negative direction. This means that the vector is pointing in the opposite direction of the positive direction, which is usually defined as counterclockwise.

5. How can the direction of a vector be changed?

The direction of a vector can be changed by adding or subtracting other vectors to it. This is known as vector addition or subtraction, which can result in a new vector with a different direction. Additionally, the direction of a vector can also be changed by rotating it around a fixed point or axis.

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