Solving Subtracting Vectors Homework Problem

[Balsam]

Homework Statement

3 m/s [E] - 5 m/s [N]

Homework Equations

I don't know how to solve this- is this done algebraically or another way?

The Attempt at a Solution

I have no clue. Would this be different if the directions were not like this--- 3 m/s 50° east - 5m/s 67° North. I have no idea how to figure this out.

 

[RJLiberator]

Based on your question you have 3m/s E. This leads me to believe that the vector is due east. Likewise with the 5m/s north.

Do you recall how to add vectors?

Subtracting is merely A-B = A+(-B).

 

[Balsam]

Based on your question you have 3m/s E. This leads me to believe that the vector is due east. Likewise with the 5m/s north.

Do you recall how to add vectors?

Subtracting is merely A-B = A+(-B).

Is this a graphical or algebraic method? I don't remember how to add vectors, either.

 

[RJLiberator]

If you look at it graphically, we have 3 m/s due east, meaning right on the x-axis. x=3
We have 5 m/s due north, meaning y = 5

A+(-B) would mean you flip B, and then place the tail of B to the head of A. Now you have a right triangle, what can do you with a right triangle to find the missing values?

 

[Chestermiller]

Suppose you had a vector 3i - 5j , where i and j are unit vectors in the x and y directions. Would you know how to find the magnitude of this vector?

 

[Balsam]

If you look at it graphically, we have 3 m/s due east, meaning right on the x-axis. x=3
We have 5 m/s due north, meaning y = 5

A+(-B) would mean you flip B, and then place the tail of B to the head of A. Now you have a right triangle, what can do you with a right triangle to find the missing values?

You find the length of the hypotenuse and acute angle? What would you do if the 2 vectors didn't form a right triangle- you wouldn't be able to use pythagorean theorem to find the hypotenuse length

 

[RJLiberator]

In this case you could break up the actual components of each vector.

If vector a was 5 m/s and at 45 degrees north of east, then you have an angle and the hypotenuse.
You should know/recall from trig that cos(angle) = a/h so h*cos(angle) = a
Now you have the velocity of that vector in the due east direction. You can do the same with sin to find the velocity in the due north direction.

Once you break up the components of the vectors, you can add/subtract the components as necessary and then pull everything back together.

 

[Balsam]

In this case you could break up the actual components of each vector.

If vector a was 5 m/s and at 45 degrees north of east, then you have an angle and the hypotenuse.
You should know/recall from trig that cos(angle) = a/h so h*cos(angle) = a
Now you have the velocity of that vector in the due east direction. You can do the same with sin to find the velocity in the due north direction.

Once you break up the components of the vectors, you can add/subtract the components as necessary and then pull everything back together.

Thank you!