Find Resultant Displacement Vector in Unit Vector Notation

In summary: Good, that looks fine. BTW, what is it about the question that makes you think it's a "discussion"?Good, that looks fine. BTW, what is it about the question that makes you think it's a "discussion"?The fact that it says "The Attempt at a Solution" and "Homework Equations" makes it seem like a discussion. Plus the fact that it says "I think" and "Did I do all the trig stuff right?" makes it seem like a discussion.The fact that it says "The Attempt at a Solution" and "Homework Equations" makes it seem like a discussion. Plus the fact that it says "I think" and "Did I do all the trig
  • #1
jeff12
40
2
1. The problem statement, all variables and given/known dat

A car travels 20 mi at 60 degrees north of west, then 35 mi at 45 degrees north of east.

Express each displacement vector in unit vector notation. Take the +x-axis due east and the +y-axis due north. Use the component method to obtain the resultant displacement vector in unit vector notation.

Homework Equations


[tex]\ A = \ A_x i + \ A_y j[/tex]
[tex]\ B = \ B_x i + \ B_y j[/tex]
[tex]\ C = \ C_x i + \ C_y j[/tex]

The Attempt at a Solution



A= -20sin60i+35cos60j
B=35cos45i+20sin45
C=?

How do I find C, since no lengths were given. Did I do A and B correctly?
 
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  • #2
jeff12 said:
1. The problem statement, all variables and given/known dat

A car travels 20 mi at 60 degrees north of west, then 35 mi at 45 degrees north of east.

Express each displacement vector in unit vector notation. Take the +x-axis due east and the +y-axis due north. Use the component method to obtain the resultant displacement vector in unit vector notation.

Homework Equations


[tex]\ A = \ A_x i + \ A_y j[/tex]
[tex]\ B = \ B_x i + \ B_y j[/tex]
[tex]\ C = \ C_x i + \ C_y j[/tex]

The Attempt at a Solution



A= -20sin60i+35cos60j=7.5
B=35cos45i+20sin45=10.6
C=?

How do I find C, since no lengths were given. Did I do A and B correctly?

The solutions look incorrect so far. For A, both the x and y components involve the 20 mile displacement, and for B, both the x and y components involve the 35 mile displacement. Also, can you show a sketch that shows how you determine when to use the cos() and when to use the sin() to find the components on the x & y axes?

Finally, you are indeed given the lengths of 20 and 35 miles, so you will be able to calculate the sum vector C = A + B. :smile:
 
  • #3
jeff12 said:
A= -20sin60i+35cos60j=7.5

Also, you can't reduce the vector components down to a single scalar number "7.5". Vectors have a magnitude and a displacement, and can be expressed as that two number combination, or in rectangular component form. Are you familiar with how to convert between those two forms?
 
  • #4
berkeman said:
Also, you can't reduce the vector components down to a single scalar number "7.5". Vectors have a magnitude and a displacement, and can be expressed as that two number combination, or in rectangular component form. Are you familiar with how to convert between those two forms?

Oh yes you are right. But are the sin and cos right?

I do not know to convert to rectangular component form. I think we are suppose to put it in unit vector notation.
 

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  • #6
berkeman said:
No. I mentioned what was wrong in my first reply.

And for converting between polar and rectangular formats of vectors, use this type of figure:

http://www.teacherschoice.com.au/maths_library/coordinates/polar_7.gif
polar_7.gif

Okay I redid everything and found:

A=20cos60i+20sin60j
B=35cos45i+35sin45j
C=(-20cos60+35cos45)i+(20sin60+35sin45)j

Did I do all the trig stuff right? I am suppose to find the magnitude too which I found was 97.62.
 
  • #7
Bump.
 
  • #8
jeff12 said:
A=20cos60i+20sin60j

I believe there is a sign error in this equation... :smile:
 
  • #9
jeff12 said:
A car travels 20 mi at 60 degrees north of west,

Would it make any difference in your solution if the direction of ##\mathrm{\vec{A}}## were instead 60° north of east?
 
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  • #10
berkeman said:
I believe there is a sign error in this equation... :smile:

Oh I fixed it. I think I understand everything now. But I want to make sure, C is the resultant right? Because C is technically R.

Mister T said:
Would it make any difference in your solution if the direction of ##\mathrm{\vec{A}}## were instead 60° north of east?

Yes because that would change the problem entirely.
 
  • #11
jeff12 said:
Okay I redid everything and found:

A=20cos60i+20sin60j
B=35cos45i+35sin45j
C=(-20cos60+35cos45)i+(20sin60+35sin45)j

Did I do all the trig stuff right? I am suppose to find the magnitude too which I found was 97.62.

berkeman said:
I believe there is a sign error in this equation... :smile:

jeff12 said:
Oh I fixed it. I think I understand everything now. But I want to make sure, C is the resultant right? Because C is technically R..

Could you show your work for the final answer for C? Thanks, that will help us answer...
 
  • #12
jeff12 said:
Yes because that would change the problem entirely.

He's just seconding my comment about the sign error... :smile:
 
  • #13
jeff12 said:
But I want to make sure, C is the resultant right? Because C is technically R.

The problem statement doesn't specify. It was you who chose ##\mathrm{\vec{A}}## to represent one of the two given vectors and ##\mathrm{\vec{B}}## the other. Likewise you are free to choose ##\mathrm{\vec{C}}## or ##\mathrm{\vec{R}}## as their sum.

If ##\mathrm{\vec{C}}## is your choice, then you'd write ##\mathrm{\vec{A}+\vec{B}=\vec C}##.

If ##\mathrm{\vec{R}}## is your choice, then you'd write ##\mathrm{\vec{A}+\vec{B}=\vec R}##.

The term "resultant" has the same meaning as "vector sum".
 
  • #14
jeff12 said:
I am suppose to find the magnitude too

No, you are not supposed to find the magnitude. You are supposed to express the result in unit vector notation.
 
  • #15
berkeman said:
Could you show your work for the final answer for C? Thanks, that will help us answer...

There was no C. I figured that out. I was confused because in his example he kept saying R=A+B+C. But since I was only given two vectors it would be R=A+B. I found R=14.75i+42.07j. The magnitude is 44.58 mi at [itex]70.68^o[/itex].
 

Related to Find Resultant Displacement Vector in Unit Vector Notation

1. What is the formula for finding the resultant displacement vector in unit vector notation?

The formula for finding the resultant displacement vector in unit vector notation is: R = √(Rx^2 + Ry^2 + Rz^2), where Rx, Ry, and Rz are the x, y, and z components of the vector, respectively.

2. How do you find the unit vector notation of a displacement vector?

To find the unit vector notation of a displacement vector, divide each component of the vector (x, y, and z) by the magnitude of the vector. The resulting vector will have a magnitude of 1 and represent the direction of the original vector.

3. Can a displacement vector be represented in both magnitude and direction?

Yes, a displacement vector can be represented in both magnitude and direction. The magnitude of the vector represents the length of the displacement, while the direction represents the orientation of the displacement.

4. How do you find the resultant displacement vector when given multiple vectors in unit vector notation?

To find the resultant displacement vector when given multiple vectors in unit vector notation, add the x, y, and z components of each vector together. Then, use the formula R = √(Rx^2 + Ry^2 + Rz^2) to find the magnitude of the resultant vector. Lastly, divide each component by the magnitude to find the unit vector notation.

5. Can the resultant displacement vector be negative?

Yes, the resultant displacement vector can be negative. The negative sign indicates the direction of the vector, so a negative resultant displacement vector would be in the opposite direction of a positive resultant displacement vector with the same magnitude.

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