Find Resultant Displacement Vector in Unit Vector Notation

[jeff12]

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

$$\ A = \ A_x i + \ A_y j$$
$$\ B = \ B_x i + \ B_y j$$
$$\ C = \ C_x i + \ C_y j$$

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?

 

[berkeman]

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

$$\ A = \ A_x i + \ A_y j$$
$$\ B = \ B_x i + \ B_y j$$
$$\ C = \ C_x i + \ C_y j$$

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.

 

[berkeman]

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?

 

[jeff12]

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.

 

[berkeman]

But are the sin and cos right?

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

 

[jeff12]

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

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.

 

[jeff12]

Bump.

 

[berkeman]

A=20cos60i+20sin60j

I believe there is a sign error in this equation...

 

[Mister T]

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?

 

[jeff12]

I believe there is a sign error in this equation...

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.

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.

 

[berkeman]

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.

I believe there is a sign error in this equation...

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...

 

[berkeman]

Yes because that would change the problem entirely.

He's just seconding my comment about the sign error...

 

[Mister T]

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".

 

[Mister T]

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.

 

[jeff12]

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 $70.68^o$.