Find X & Y Head of Vectors: Speed & Angle Given

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In summary, the sailboat is experiencing a force of 175 N at an angle of 20 degrees west and a second force of 225 N at an angle of 20 degrees south.
  • #1
Laythen
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Hi. I have a problem I need to figure out, I don't need help doing the problem so much as I need help with just figuring out an equation.

A sailboat traveling at a speed of 40km/h in a direction of N 20 degrees E encounters a current that causes it to veer off-course at a speed of 10m/h in a direction of E 20 degrees N. Determine the sailboat's actual speed and direction.


I need help figuring out how to find the x and y head of my vector going at 40km/h at an angle of N 20 degrees E and then for the head of my vector going at 10m/h in a direction of E 20 degrees N.

I am given only the information above and need to find how long to make my vector V and my vector U. I need to know how to get the head of vector V and vector U given only their speed and angle.

If you can help me with the equation to get (x,y) for the heads, I can then draw the vectors and go on to do my parallelogram and find a solution to the problem. Please help me and let me know what the equation to finding the head of a vector is with only speed and angle. Thank you.
 
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  • #2
Hi Laythen welcome to PF,

they have given you a question that confuses me haha. Do they mean to say that your final velocity is 10m/h in a direction of E 20 degrees N or that the current's velocity is 10m/h in a direction of E 20 degrees N?
 
  • #3
Also does "N 20 degrees E" mean 20 degrees north of due east? (That problem was not composed by any sailor!)
 
  • #4
A vector of length h at angle [itex]\theta[/itex] has x component [itex]h cos(\theta)[/itex] and y component [itex]h sin(\theta)[/itex].

Since they ask for the "sailboat's actual speed and direction" the second vector is the current, not the sailboats new vector.
 
  • #5
obafgkmrns said:
Also does "N 20 degrees E" mean 20 degrees north of due east? (That problem was not composed by any sailor!)

Yes that is it

HallsofIvy said:
A vector of length h at angle [itex]\theta[/itex] has x component [itex]h cos(\theta)[/itex] and y component [itex]h sin(\theta)[/itex].

Since they ask for the "sailboat's actual speed and direction" the second vector is the current, not the sailboats new vector.

Hi. I private messaged you this problem on Mathhelp yesterday but it seems to be down. Yes you gave me that equation for a previous problem I had and I thank you very much, it allowed me to continue in my book.

I don't know how to apply it to this problem though.

A sailboat traveling at a speed of 40km/h in a direction of N 20 degrees E encounters a current that causes it to veer off-course at a speed of 10m/h in a direction of E 20 degrees N. Determine the sailboat's actual speed and direction.

Let's take the first vector and call it V.

Vector V is traveling at a speed of 40km/h in a direction of N 20 degrees E.

With that information, is it possible to graph a vector and how long will it be? I need to know how long to make it and where its head is. I can see looking at the answer sheet that the coordinates of the head are (7,13). How do I get this from only knowing the speed and angle?

I also know that north is 90 degrees and so 20 degrees east from north is 90-20 = 70 degrees.

So I have this information and need to make a vector

N 20 degrees E (So 70 degrees) going at 40km/h.

Any help appreciated. Thank you
 
  • #6
The problem seems to be asking you to add two vectors, i.e., to add the 40km/h "course" north component "course" to the 10km/h "current" north component, and ditto the east components. The convert the resulting vector to speed and heading.

(Real sailboats don't act that way. A change in current results in a change in apparent wind, which results in a change in sail set, which results in a change in speed with respect to the water. Also boat travel in knots, not km/h!)
 
  • #7
obafgkmrns said:
Also boat travel in knots, not km/h!)

And on a similar note, one of the units speeds was quoted in mph while the other one kmph.
 
  • #8
Yes I know, the book isn't entirely accurate. I need to know how to solve the first question I asked but let me give you guys another question that comes after it. It is possible to solve these questions without graphing them, but I will need to know how to graph them.A force of F1 of 175 N at an angle of direction N 45 W and a second force F2 of 225 N at an angle of direction of W 20 S are exerted on a concrete block in order to move it along a horizontal plane. Determine the norm and measure of the angle of direction of the resultant of these two forces.

OK. Let's say F1 is vector.. F1. Makes sense.

Here's what we know about F1:

Force-175 N
Angle of direction - N 45 degrees W

N 45 degrees W means 90 + 45 = 135 degrees

Ok. So how long will I make vector F1 with the information above? From the answer sheet, this vector goes from the origin until its head reaches point (-4,4)

What equation do I use to get the head (-4,4) with the information I have? (force 175 N, 135 degrees)

Any help. Thanks.
 
  • #9
Final bump, if anyone knows the equation to either question, please help. Thank you
 

FAQ: Find X & Y Head of Vectors: Speed & Angle Given

What is the formula for finding the head of a vector given its speed and angle?

The formula for finding the head of a vector given its speed and angle is: head = speed * cos(angle)

How do I find the speed of a vector if I know its head and angle?

To find the speed of a vector if you know its head and angle, you can use the formula: speed = head / cos(angle)

Can I use this formula for both 2D and 3D vectors?

Yes, this formula can be used for both 2D and 3D vectors as long as the angle is measured from the x-axis.

Is there a specific unit for the speed of a vector?

The unit for the speed of a vector depends on the unit used for distance and time. For example, if distance is measured in meters and time is measured in seconds, then the unit for speed would be meters per second (m/s).

Can I find the angle of a vector if I know its head and speed?

Yes, you can find the angle of a vector if you know its head and speed by using the formula: angle = arccos(head / speed)

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